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feat: add solutions to lc problem: No.1092
No.1092.Shortest Common Supersequence
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‎solution/1000-1099/1092.Shortest Common Supersequence/README.md

Lines changed: 175 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -35,22 +35,196 @@ str2 = "cab" 是 "cabac" 的一个子串，因为我们可
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划 + 构造**
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我们先用动态规划求出两个字符串的最长公共子序列，然后根据最长公共子序列构造出最短公共超序列。
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定义 $f[i][j]$ 表示字符串 $str1$ 的前 $i$ 个字符和字符串 $str2$ 的前 $j$ 个字符的最长公共子序列的长度。状态转移方程如下：
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$$
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f[i][j] = \left\{\begin{matrix}
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0 & i = 0 \text{ or } j = 0 \\
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f[i - 1][j - 1] + 1 & str1[i - 1] = str2[j - 1] \\
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\max(f[i - 1][j], f[i][j - 1]) & str1[i - 1] \neq str2[j - 1]
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\end{matrix}\right.
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$$
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接下来我们基于 $f[i][j]$ 构造出最短公共超序列。
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用双指针 $i$ 和 $j$ 分别指向字符串 $str1$ 和 $str2$ 的末尾，然后从后往前遍历，每次比较 $str1[i]$ 和 $str2[j]$ 的值，如果 $str1[i] = str2[j]$，则将 $str1[i]$ 或 $str2[j]$ 中的任意一个字符加入到最短公共超序列的末尾，然后 $i$ 和 $j$ 同时减 1；如果 $str1[i] \neq str2[j]$，则将 $f[i][j]$ 与 $f[i - 1][j]$ 和 $f[i][j - 1]$ 中的最大值进行比较，如果 $f[i][j] = f[i - 1][j]$，则将 $str1[i]$ 加入到最短公共超序列的末尾，然后 $i$ 减 1；如果 $f[i][j] = f[i][j - 1]$，则将 $str2[j]$ 加入到最短公共超序列的末尾，然后 $j$ 减 1。重复上述操作，直到 $i = 0$ 或 $j = 0$，然后将剩余的字符串加入到最短公共超序列的末尾即可。
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时间复杂度 $O(m\times n)$，空间复杂度 $O(m\times n)$。其中 $m$ 和 $n$ 分别是字符串 $str1$ 和 $str2$ 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
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m, n = len(str1), len(str2)
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f = [[0] * (n + 1) for _ in range(m + 1)]
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for i in range(1, m + 1):
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for j in range(1, n + 1):
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if str1[i - 1] == str2[j - 1]:
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f[i][j] = f[i - 1][j - 1] + 1
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else:
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f[i][j] = max(f[i - 1][j], f[i][j - 1])
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ans = []
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i, j = m, n
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while i or j:
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if i == 0:
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j -= 1
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ans.append(str2[j])
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elif j == 0:
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i -= 1
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ans.append(str1[i])
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else:
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if f[i][j] == f[i - 1][j]:
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i -= 1
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ans.append(str1[i])
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elif f[i][j] == f[i][j - 1]:
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j -= 1
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ans.append(str2[j])
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else:
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i, j = i - 1, j - 1
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ans.append(str1[i])
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return ''.join(ans[::-1])
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String shortestCommonSupersequence(String str1, String str2) {
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int m = str1.length(), n = str2.length();
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int[][] f = new int[m + 1][n + 1];
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
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f[i][j] = f[i - 1][j - 1] + 1;
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} else {
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f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
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}
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}
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}
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int i = m, j = n;
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StringBuilder ans = new StringBuilder();
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while (i > 0 || j > 0) {
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if (i == 0) {
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ans.append(str2.charAt(--j));
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} else if (j == 0) {
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ans.append(str1.charAt(--i));
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} else {
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if (f[i][j] == f[i - 1][j]) {
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ans.append(str1.charAt(--i));
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} else if (f[i][j] == f[i][j - 1]) {
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ans.append(str2.charAt(--j));
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} else {
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ans.append(str1.charAt(--i));
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--j;
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}
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}
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}
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return ans.reverse().toString();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string shortestCommonSupersequence(string str1, string str2) {
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int m = str1.size(), n = str2.size();
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vector<vector<int>> f(m + 1, vector<int>(n + 1));
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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if (str1[i - 1] == str2[j - 1])
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f[i][j] = f[i - 1][j - 1] + 1;
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else
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f[i][j] = max(f[i - 1][j], f[i][j - 1]);
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}
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}
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int i = m, j = n;
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string ans;
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while (i || j) {
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if (i == 0)
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ans += str2[--j];
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else if (j == 0)
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ans += str1[--i];
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else {
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if (f[i][j] == f[i - 1][j])
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ans += str1[--i];
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else if (f[i][j] == f[i][j - 1])
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ans += str2[--j];
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else
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ans += str1[--i], --j;
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}
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}
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reverse(ans.begin(), ans.end());
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return ans;
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}
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};
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```
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### **Go**
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```go
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func shortestCommonSupersequence(str1 string, str2 string) string {
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m, n := len(str1), len(str2)
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f := make([][]int, m+1)
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for i := range f {
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f[i] = make([]int, n+1)
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}
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
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if str1[i-1] == str2[j-1] {
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f[i][j] = f[i-1][j-1] + 1
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} else {
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f[i][j] = max(f[i-1][j], f[i][j-1])
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}
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}
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}
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ans := []byte{}
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i, j := m, n
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for i > 0 || j > 0 {
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if i == 0 {
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j--
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ans = append(ans, str2[j])
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} else if j == 0 {
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i--
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ans = append(ans, str1[i])
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} else {
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if f[i][j] == f[i-1][j] {
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i--
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ans = append(ans, str1[i])
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} else if f[i][j] == f[i][j-1] {
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j--
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ans = append(ans, str2[j])
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} else {
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i, j = i-1, j-1
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ans = append(ans, str1[i])
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}
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}
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}
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for i, j = 0, len(ans)-1; i < j; i, j = i+1, j-1 {
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ans[i], ans[j] = ans[j], ans[i]
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}
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return string(ans)
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1000-1099/1092.Shortest Common Supersequence/README_EN.md

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@@ -42,13 +42,167 @@ The answer provided is the shortest such string that satisfies these properties.
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### **Python3**
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```python
45-
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class Solution:
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def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
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m, n = len(str1), len(str2)
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f = [[0] * (n + 1) for _ in range(m + 1)]
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for i in range(1, m + 1):
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for j in range(1, n + 1):
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if str1[i - 1] == str2[j - 1]:
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f[i][j] = f[i - 1][j - 1] + 1
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else:
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f[i][j] = max(f[i - 1][j], f[i][j - 1])
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ans = []
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i, j = m, n
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while i or j:
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if i == 0:
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j -= 1
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ans.append(str2[j])
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elif j == 0:
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i -= 1
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ans.append(str1[i])
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else:
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if f[i][j] == f[i - 1][j]:
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i -= 1
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ans.append(str1[i])
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elif f[i][j] == f[i][j - 1]:
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j -= 1
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ans.append(str2[j])
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else:
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i, j = i - 1, j - 1
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ans.append(str1[i])
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return ''.join(ans[::-1])
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```
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### **Java**
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```java
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class Solution {
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public String shortestCommonSupersequence(String str1, String str2) {
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int m = str1.length(), n = str2.length();
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int[][] f = new int[m + 1][n + 1];
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
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f[i][j] = f[i - 1][j - 1] + 1;
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} else {
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f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
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}
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}
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}
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int i = m, j = n;
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StringBuilder ans = new StringBuilder();
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while (i > 0 || j > 0) {
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if (i == 0) {
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ans.append(str2.charAt(--j));
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} else if (j == 0) {
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ans.append(str1.charAt(--i));
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} else {
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if (f[i][j] == f[i - 1][j]) {
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ans.append(str1.charAt(--i));
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} else if (f[i][j] == f[i][j - 1]) {
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ans.append(str2.charAt(--j));
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} else {
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ans.append(str1.charAt(--i));
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--j;
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}
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}
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}
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return ans.reverse().toString();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string shortestCommonSupersequence(string str1, string str2) {
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int m = str1.size(), n = str2.size();
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vector<vector<int>> f(m + 1, vector<int>(n + 1));
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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if (str1[i - 1] == str2[j - 1])
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f[i][j] = f[i - 1][j - 1] + 1;
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else
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f[i][j] = max(f[i - 1][j], f[i][j - 1]);
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}
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}
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int i = m, j = n;
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string ans;
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while (i || j) {
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if (i == 0)
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ans += str2[--j];
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else if (j == 0)
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ans += str1[--i];
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else {
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if (f[i][j] == f[i - 1][j])
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ans += str1[--i];
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else if (f[i][j] == f[i][j - 1])
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ans += str2[--j];
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else
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ans += str1[--i], --j;
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}
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}
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reverse(ans.begin(), ans.end());
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return ans;
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}
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};
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```
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### **Go**
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```go
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func shortestCommonSupersequence(str1 string, str2 string) string {
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m, n := len(str1), len(str2)
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f := make([][]int, m+1)
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for i := range f {
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f[i] = make([]int, n+1)
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}
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
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if str1[i-1] == str2[j-1] {
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f[i][j] = f[i-1][j-1] + 1
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} else {
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f[i][j] = max(f[i-1][j], f[i][j-1])
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}
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}
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}
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ans := []byte{}
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i, j := m, n
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for i > 0 || j > 0 {
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if i == 0 {
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j--
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ans = append(ans, str2[j])
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} else if j == 0 {
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i--
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ans = append(ans, str1[i])
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} else {
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if f[i][j] == f[i-1][j] {
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i--
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ans = append(ans, str1[i])
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} else if f[i][j] == f[i][j-1] {
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j--
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ans = append(ans, str2[j])
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} else {
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i, j = i-1, j-1
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ans = append(ans, str1[i])
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}
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}
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}
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for i, j = 0, len(ans)-1; i < j; i, j = i+1, j-1 {
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ans[i], ans[j] = ans[j], ans[i]
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}
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return string(ans)
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}
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func max(a, b int) int {
201+
if a > b {
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return a
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}
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return b
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}
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```
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### **...**

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