solve 567.字符串的排列

Author: andavid

zh/567.字符串的排列.java

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+/*
+ * @lc app=leetcode.cn id=567 lang=java
+ *
+ * [567] 字符串的排列
+ *
+ * https://leetcode-cn.com/problems/permutation-in-string/description/
+ *
+ * algorithms
+ * Medium (35.79%)
+ * Likes:    151
+ * Dislikes: 0
+ * Total Accepted:    33.3K
+ * Total Submissions: 91.4K
+ * Testcase Example:  '"ab"\n"eidbaooo"'
+ *
+ * 给定两个字符串 s1 和 s2，写一个函数来判断 s2 是否包含 s1 的排列。
+ *
+ * 换句话说，第一个字符串的排列之一是第二个字符串的子串。
+ *
+ * 示例1:
+ *
+ *
+ * 输入: s1 = "ab" s2 = "eidbaooo"
+ * 输出: True
+ * 解释: s2 包含 s1 的排列之一 ("ba").
+ *
+ *
+ *
+ *
+ * 示例2:
+ *
+ *
+ * 输入: s1= "ab" s2 = "eidboaoo"
+ * 输出: False
+ *
+ *
+ *
+ *
+ * 注意：
+ *
+ *
+ * 输入的字符串只包含小写字母
+ * 两个字符串的长度都在 [1, 10,000] 之间
+ *
+ *
+ */
+
+// @lc code=start
+class Solution {
+  public boolean checkInclusion(String s1, String s2) {
+    int[] need = new int[128];
+    int[] window = new int[128];
+    for (char ch : s1.toCharArray()) {
+      need[ch]++;
+    }
+
+    int left = 0, right = 0;
+    int count = 0;
+
+    while (right < s2.length()) {
+      char ch = s2.charAt(right);
+      right++;
+      if (need[ch] > 0) {
+        window[ch]++;
+        if (window[ch] <= need[ch]) {
+          count++;
+        }
+      }
+
+      while (right - left >= s1.length()) {
+        if (count == s1.length()) {
+          return true;
+        }
+
+        ch = s2.charAt(left);
+        left++;
+        if (need[ch] > 0) {
+          if (window[ch] <= need[ch]) {
+            count--;
+          }
+          window[ch]--;
+        }
+      }
+    }
+
+    return false;
+  }
+}
+// @lc code=end
+