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feat: add solutions to lc problem: No.0567
No.0567.Permutation in String
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‎solution/0500-0599/0567.Permutation in String/README.md

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@@ -40,6 +40,22 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：滑动窗口**
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我们观察发现，题目实际上等价于判断字符串 $s2$ 中是否存在窗口大小为 $n$，且窗口内的字符及其个数与字符串 $s1$ 相同的子串。
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因此，我们先用哈希表或数组 $cnt1$ 统计字符串 $s1$ 中每个字符出现的次数，然后遍历字符串 $s2$，维护一个窗口大小为 $n$ 的滑动窗口，用哈希表或数组 $cnt2$ 统计窗口内每个字符出现的次数，当 $cnt1 = cnt2$ 时，说明窗口内的字符及其个数与字符串 $s1$ 相同，返回 `true` 即可。
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否则，遍历结束后，返回 `false`。
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时间复杂度 $(n + m \times C)$，空间复杂度 $O(C)$。其中 $n$ 和 $m$ 分别为字符串 $s1$ 和 $s2$ 的长度；而 $C$ 为字符集的大小，本题中 $C=26$。
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**方法二：滑动窗口优化**
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在方法一中，我们每次加入和移除一个字符时，都需要比较两个哈希表或数组，时间复杂度较高。我们可以维护一个变量 $diff$，表示两个大小为 $n$ 的字符串中，有多少种字符出现的个数不同。当 $diff=0$ 时，说明两个字符串中的字符个数相同。
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时间复杂度 $O(n + m)$，空间复杂度 $O(C)$。其中 $n$ 和 $m$ 分别为字符串 $s1$ 和 $s2$ 的长度；而 $C$ 为字符集的大小，本题中 $C=26$。
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<!-- tabs:start -->
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### **Python3**
@@ -49,30 +65,49 @@
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```python
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class Solution:
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def checkInclusion(self, s1: str, s2: str) -> bool:
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need, window = {}, {}
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validate, left, right = 0, 0, 0
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for c in s1:
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window[c] = 0
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if c in need:
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need[c] += 1
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else:
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need[c] = 1
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for right in range(len(s2)):
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c = s2[right]
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if c in need:
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window[c] += 1
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if window[c] == need[c]:
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validate += 1
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while right - left + 1 >= len(s1):
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if validate == len(need):
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return True
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d = s2[left]
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left += 1
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if d in need:
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if window[d] == need[d]:
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validate -= 1
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window[d] -= 1
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n = len(s1)
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cnt1 = Counter(s1)
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cnt2 = Counter(s2[:n])
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if cnt1 == cnt2:
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return True
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for i in range(n, len(s2)):
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cnt2[s2[i]] += 1
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cnt2[s2[i - n]] -= 1
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if cnt1 == cnt2:
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return True
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return False
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```
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```python
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class Solution:
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def checkInclusion(self, s1: str, s2: str) -> bool:
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n, m = len(s1), len(s2)
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if n > m:
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return False
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cnt = Counter()
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for a, b in zip(s1, s2):
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cnt[a] -= 1
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cnt[b] += 1
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diff = sum(x != 0 for x in cnt.values())
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if diff == 0:
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return True
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for i in range(n, m):
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a, b = s2[i - n], s2[i]
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if cnt[b] == 0:
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diff += 1
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cnt[b] += 1
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if cnt[b] == 0:
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diff -= 1
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if cnt[a] == 0:
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diff += 1
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cnt[a] -= 1
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if cnt[a] == 0:
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diff -= 1
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if diff == 0:
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return True
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return False
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```
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@@ -81,7 +116,226 @@ class Solution:
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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 boolean checkInclusion(String s1, String s2) {
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int n = s1.length();
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int m = s2.length();
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if (n > m) {
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return false;
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}
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int[] cnt1 = new int[26];
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int[] cnt2 = new int[26];
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for (int i = 0; i < n; ++i) {
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++cnt1[s1.charAt(i) - 'a'];
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++cnt2[s2.charAt(i) - 'a'];
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}
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if (Arrays.equals(cnt1, cnt2)) {
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return true;
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}
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for (int i = n; i < m; ++i) {
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++cnt2[s2.charAt(i) - 'a'];
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--cnt2[s2.charAt(i - n) - 'a'];
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if (Arrays.equals(cnt1, cnt2)) {
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return true;
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}
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}
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return false;
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}
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}
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```
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```java
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class Solution {
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public boolean checkInclusion(String s1, String s2) {
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int n = s1.length();
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int m = s2.length();
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if (n > m) {
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return false;
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}
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int[] cnt = new int[26];
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for (int i = 0; i < n; ++i) {
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--cnt[s1.charAt(i) - 'a'];
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++cnt[s2.charAt(i) - 'a'];
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}
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int diff = 0;
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for (int x : cnt) {
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if (x != 0) {
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++diff;
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}
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}
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if (diff == 0) {
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return true;
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}
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for (int i = n; i < m; ++i) {
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int a = s2.charAt(i - n) - 'a';
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int b = s2.charAt(i) - 'a';
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if (cnt[b] == 0) {
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++diff;
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}
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if (++cnt[b] == 0) {
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--diff;
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}
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if (cnt[a] == 0) {
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++diff;
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}
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if (--cnt[a] == 0) {
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--diff;
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}
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if (diff == 0) {
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return true;
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}
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}
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return false;
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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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bool checkInclusion(string s1, string s2) {
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int n = s1.size(), m = s2.size();
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if (n > m) {
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return false;
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}
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vector<int> cnt1(26), cnt2(26);
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for (int i = 0; i < n; ++i) {
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++cnt1[s1[i] - 'a'];
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++cnt2[s2[i] - 'a'];
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}
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if (cnt1 == cnt2) {
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return true;
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}
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for (int i = n; i < m; ++i) {
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++cnt2[s2[i] - 'a'];
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--cnt2[s2[i - n] - 'a'];
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if (cnt1 == cnt2) {
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return true;
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}
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}
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return false;
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}
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};
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```
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```cpp
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class Solution {
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public:
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bool checkInclusion(string s1, string s2) {
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int n = s1.size(), m = s2.size();
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if (n > m) {
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return false;
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}
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vector<int> cnt(26);
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for (int i = 0; i < n; ++i) {
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--cnt[s1[i] - 'a'];
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++cnt[s2[i] - 'a'];
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}
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int diff = 0;
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for (int x : cnt) {
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if (x) {
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++diff;
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}
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}
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if (diff == 0) {
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return true;
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}
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for (int i = n; i < m; ++i) {
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int a = s2[i - n] - 'a';
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int b = s2[i] - 'a';
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if (cnt[b] == 0) {
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++diff;
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}
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if (++cnt[b] == 0) {
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--diff;
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}
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if (cnt[a] == 0) {
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++diff;
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}
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if (--cnt[a] == 0) {
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--diff;
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}
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if (diff == 0) {
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return true;
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}
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}
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return false;
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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 checkInclusion(s1 string, s2 string) bool {
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n, m := len(s1), len(s2)
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if n > m {
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return false
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}
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cnt1 := [26]int{}
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cnt2 := [26]int{}
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for i := range s1 {
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cnt1[s1[i]-'a']++
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cnt2[s2[i]-'a']++
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}
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if cnt1 == cnt2 {
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return true
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}
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for i := n; i < m; i++ {
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cnt2[s2[i]-'a']++
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cnt2[s2[i-n]-'a']--
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if cnt1 == cnt2 {
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return true
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}
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}
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return false
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}
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```
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```go
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func checkInclusion(s1 string, s2 string) bool {
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n, m := len(s1), len(s2)
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if n > m {
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return false
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}
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cnt := [26]int{}
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for i := range s1 {
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cnt[s1[i]-'a']--
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cnt[s2[i]-'a']++
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}
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diff := 0
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for _, x := range cnt {
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if x != 0 {
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diff++
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}
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}
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if diff == 0 {
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return true
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}
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for i := n; i < m; i++ {
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a, b := s2[i-n]-'a', s2[i]-'a'
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if cnt[b] == 0 {
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diff++
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}
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cnt[b]++
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if cnt[b] == 0 {
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diff--
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}
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if cnt[a] == 0 {
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diff++
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}
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cnt[a]--
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if cnt[a] == 0 {
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diff--
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}
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if diff == 0 {
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return true
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}
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}
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return false
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}
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```
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### **TypeScript**

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