131. 分割回文串

Author: w2xi

README.md

@@ -72,6 +72,7 @@
 |124|[二叉树中的最大路径和](https://leetcode.cn/problems/binary-tree-maximum-path-sum/)|[JavaScript](./algorithms/binary-tree-maximum-path-sum.js)|Hard|
 |125|[验证回文串](https://leetcode.cn/problems/valid-palindrome/)|[JavaScript](./algorithms/valid-palindrome.js)|Easy|
 |129|[求根节点到叶节点数字之和](https://leetcode.cn/problems/sum-root-to-leaf-numbers/)|[JavaScript](./algorithms/sum-root-to-leaf-numbers.js)|Medium|
+|131|[分割回文串](https://leetcode.cn/problems/palindrome-partitioning/)|[JavaScript](./algorithms/palindrome-partitioning.js)|Medium|
 |136|[只出现一次的数字](https://leetcode-cn.com/problems/single-number/)|[JavaScript](./algorithms/single-number.js)|Easy|
 |141|[环形链表](https://leetcode-cn.com/problems/linked-list-cycle/)|[JavaScript](./algorithms/linked-list-cycle.js)|Easy|
 |142|[环形链表 II](https://leetcode.cn/problems/linked-list-cycle-ii/)|[JavaScript](./algorithms/linked-list-cycle-ii.js)|Medium|

algorithms/palindrome-partitioning.js

@@ -0,0 +1,40 @@
+/**
+ * 131. 分割回文串
+ * @param {string} s
+ * @return {string[][]}
+ */
+var partition = function (s) {
+  const result = [];
+  const path = [];
+
+  const backtracking = (s, startIndex = 0) => {
+    if (startIndex >= s.length) {
+      result.push(path.slice());
+      return;
+    }
+    for (let i = startIndex; i < s.length; i++) {
+      if (isPalindrome(s, startIndex, i)) {
+        const str = s.slice(startIndex, i + 1);
+        path.push(str);
+      } else {
+        continue;
+      }
+      backtracking(s, i + 1);
+      path.pop();
+    }
+  };
+  backtracking(s);
+
+  return result;
+};
+
+function isPalindrome(s, left, right) {
+  while (left < right) {
+    if (s[left] !== s[right]) {
+      return false;
+    }
+    left++;
+    right--;
+  }
+  return true;
+}