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feat: add solutions to lc problem: No.1868
No.1868.Product of Two Run-Length Encoded Arrays
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‎solution/1800-1899/1868.Product of Two Run-Length Encoded Arrays/README.md

Lines changed: 114 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -61,22 +61,135 @@ prodNums = [2,2,2,6,9,9]，压缩成行程编码数组 [[2,3],[6,1],[9,2]]。
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：双指针**
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我们用两个指针 $i$ 和 $j$ 分别指向两个数组的当前位置，然后开始模拟乘法的过程。
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对于当前位置 $i$ 和 $j$，我们取 $f=min(encoded1[i][1],encoded2[j][1])$，表示当前位置的乘积的频次，然后将 $v=encoded1[i][0] \times encoded2[j][0]$，表示当前位置的乘积的值。如果当前位置的乘积的值 $v$ 和上一个位置的乘积的值相同，则将当前位置的乘积的频次加到上一个位置的乘积的频次上，否则将当前位置的乘积的值和频次加到答案数组中。然后我们将 $encoded1[i][1]$ 和 $encoded2[j][1]$ 分别减去 $f$，如果 $encoded1[i][1]$ 或 $encoded2[j][1]$ 减为 $0$，则将对应的指针向后移动一位。
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最后返回答案数组即可。
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时间复杂度 $O(m + n)$，其中 $m$ 和 $n$ 分别是两个数组的长度。忽略答案数组的空间消耗，空间复杂度 $O(1)$。
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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 findRLEArray(self, encoded1: List[List[int]], encoded2: List[List[int]]) -> List[List[int]]:
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ans = []
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j = 0
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for vi, fi in encoded1:
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while fi:
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f = min(fi, encoded2[j][1])
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v = vi * encoded2[j][0]
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if ans and ans[-1][0] == v:
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ans[-1][1] += f
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else:
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ans.append([v, f])
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fi -= f
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encoded2[j][1] -= f
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if encoded2[j][1] == 0:
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j += 1
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return ans
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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 List<List<Integer>> findRLEArray(int[][] encoded1, int[][] encoded2) {
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List<List<Integer>> ans = new ArrayList<>();
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int j = 0;
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for (var e : encoded1) {
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int vi = e[0], fi = e[1];
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while (fi > 0) {
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int f = Math.min(fi, encoded2[j][1]);
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int v = vi * encoded2[j][0];
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int m = ans.size();
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if (m > 0 && ans.get(m - 1).get(0) == v) {
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ans.get(m - 1).set(1, ans.get(m - 1).get(1) + f);
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} else {
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ans.add(new ArrayList<>(List.of(v, f)));
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}
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fi -= f;
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encoded2[j][1] -= f;
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if (encoded2[j][1] == 0) {
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++j;
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}
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}
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}
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return ans;
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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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vector<vector<int>> findRLEArray(vector<vector<int>>& encoded1, vector<vector<int>>& encoded2) {
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vector<vector<int>> ans;
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int j = 0;
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for (auto& e : encoded1) {
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int vi = e[0], fi = e[1];
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while (fi) {
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int f = min(fi, encoded2[j][1]);
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int v = vi * encoded2[j][0];
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if (!ans.empty() && ans.back()[0] == v) {
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ans.back()[1] += f;
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} else {
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ans.push_back({v, f});
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}
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fi -= f;
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encoded2[j][1] -= f;
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if (encoded2[j][1] == 0) {
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++j;
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}
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}
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}
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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 findRLEArray(encoded1 [][]int, encoded2 [][]int) (ans [][]int) {
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j := 0
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for _, e := range encoded1 {
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vi, fi := e[0], e[1]
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for fi > 0 {
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f := min(fi, encoded2[j][1])
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v := vi * encoded2[j][0]
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if len(ans) > 0 && ans[len(ans)-1][0] == v {
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ans[len(ans)-1][1] += f
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} else {
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ans = append(ans, []int{v, f})
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}
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fi -= f
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encoded2[j][1] -= f
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if encoded2[j][1] == 0 {
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j++
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}
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}
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}
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return
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}
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func min(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/1800-1899/1868.Product of Two Run-Length Encoded Arrays/README_EN.md

Lines changed: 104 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -62,13 +62,116 @@ prodNums = [2,2,2,6,9,9], which is compressed into the run-length encoded array
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### **Python3**
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```python
65-
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class Solution:
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def findRLEArray(self, encoded1: List[List[int]], encoded2: List[List[int]]) -> List[List[int]]:
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ans = []
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j = 0
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for vi, fi in encoded1:
70+
while fi:
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f = min(fi, encoded2[j][1])
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v = vi * encoded2[j][0]
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if ans and ans[-1][0] == v:
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ans[-1][1] += f
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else:
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ans.append([v, f])
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fi -= f
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encoded2[j][1] -= f
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if encoded2[j][1] == 0:
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j += 1
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return ans
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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 List<List<Integer>> findRLEArray(int[][] encoded1, int[][] encoded2) {
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List<List<Integer>> ans = new ArrayList<>();
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int j = 0;
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for (var e : encoded1) {
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int vi = e[0], fi = e[1];
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while (fi > 0) {
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int f = Math.min(fi, encoded2[j][1]);
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int v = vi * encoded2[j][0];
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int m = ans.size();
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if (m > 0 && ans.get(m - 1).get(0) == v) {
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ans.get(m - 1).set(1, ans.get(m - 1).get(1) + f);
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} else {
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ans.add(new ArrayList<>(List.of(v, f)));
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}
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fi -= f;
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encoded2[j][1] -= f;
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if (encoded2[j][1] == 0) {
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++j;
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}
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}
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}
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return ans;
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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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vector<vector<int>> findRLEArray(vector<vector<int>>& encoded1, vector<vector<int>>& encoded2) {
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vector<vector<int>> ans;
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int j = 0;
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for (auto& e : encoded1) {
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int vi = e[0], fi = e[1];
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while (fi) {
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int f = min(fi, encoded2[j][1]);
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int v = vi * encoded2[j][0];
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if (!ans.empty() && ans.back()[0] == v) {
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ans.back()[1] += f;
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} else {
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ans.push_back({v, f});
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}
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fi -= f;
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encoded2[j][1] -= f;
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if (encoded2[j][1] == 0) {
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++j;
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}
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}
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}
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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 findRLEArray(encoded1 [][]int, encoded2 [][]int) (ans [][]int) {
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j := 0
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for _, e := range encoded1 {
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vi, fi := e[0], e[1]
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for fi > 0 {
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f := min(fi, encoded2[j][1])
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v := vi * encoded2[j][0]
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if len(ans) > 0 && ans[len(ans)-1][0] == v {
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ans[len(ans)-1][1] += f
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} else {
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ans = append(ans, []int{v, f})
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}
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fi -= f
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encoded2[j][1] -= f
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if encoded2[j][1] == 0 {
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j++
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}
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}
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}
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return
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}
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func min(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/1800-1899/1868.Product of Two Run-Length Encoded Arrays/Solution.cpp

Lines changed: 25 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,25 @@
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class Solution {
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public:
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vector<vector<int>> findRLEArray(vector<vector<int>>& encoded1, vector<vector<int>>& encoded2) {
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vector<vector<int>> ans;
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int j = 0;
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for (auto& e : encoded1) {
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int vi = e[0], fi = e[1];
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while (fi) {
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int f = min(fi, encoded2[j][1]);
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int v = vi * encoded2[j][0];
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if (!ans.empty() && ans.back()[0] == v) {
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ans.back()[1] += f;
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} else {
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ans.push_back({v, f});
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}
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fi -= f;
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encoded2[j][1] -= f;
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if (encoded2[j][1] == 0) {
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++j;
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}
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}
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}
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return ans;
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}
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};

‎solution/1800-1899/1868.Product of Two Run-Length Encoded Arrays/Solution.go

Lines changed: 28 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,28 @@
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func findRLEArray(encoded1 [][]int, encoded2 [][]int) (ans [][]int) {
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j := 0
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for _, e := range encoded1 {
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vi, fi := e[0], e[1]
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for fi > 0 {
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f := min(fi, encoded2[j][1])
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v := vi * encoded2[j][0]
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if len(ans) > 0 && ans[len(ans)-1][0] == v {
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ans[len(ans)-1][1] += f
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} else {
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ans = append(ans, []int{v, f})
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}
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fi -= f
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encoded2[j][1] -= f
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if encoded2[j][1] == 0 {
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j++
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}
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}
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}
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return
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}
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func min(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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}

‎solution/1800-1899/1868.Product of Two Run-Length Encoded Arrays/Solution.java

Lines changed: 25 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,25 @@
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class Solution {
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public List<List<Integer>> findRLEArray(int[][] encoded1, int[][] encoded2) {
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List<List<Integer>> ans = new ArrayList<>();
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int j = 0;
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for (var e : encoded1) {
6+
int vi = e[0], fi = e[1];
7+
while (fi > 0) {
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int f = Math.min(fi, encoded2[j][1]);
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int v = vi * encoded2[j][0];
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int m = ans.size();
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if (m > 0 && ans.get(m - 1).get(0) == v) {
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ans.get(m - 1).set(1, ans.get(m - 1).get(1) + f);
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} else {
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ans.add(new ArrayList<>(List.of(v, f)));
15+
}
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fi -= f;
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encoded2[j][1] -= f;
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if (encoded2[j][1] == 0) {
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++j;
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}
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}
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}
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return ans;
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}
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}

‎solution/1800-1899/1868.Product of Two Run-Length Encoded Arrays/Solution.py

Lines changed: 17 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,17 @@
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class Solution:
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def findRLEArray(self, encoded1: List[List[int]], encoded2: List[List[int]]) -> List[List[int]]:
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ans = []
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j = 0
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for vi, fi in encoded1:
6+
while fi:
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f = min(fi, encoded2[j][1])
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v = vi * encoded2[j][0]
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if ans and ans[-1][0] == v:
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ans[-1][1] += f
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else:
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ans.append([v, f])
13+
fi -= f
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encoded2[j][1] -= f
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if encoded2[j][1] == 0:
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j += 1
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return ans

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