feat: azl397985856#125 azl397985856#131

assets/drawio/125.valid-palindrome.drawio

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problems/125.valid-palindrome.md

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+
+## 题目地址
+
+https://leetcode.com/problems/valid-palindrome/description/
+
+## 题目描述
+
+```
+Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
+
+Note: For the purpose of this problem, we define empty string as valid palindrome.
+
+Example 1:
+
+Input: "A man, a plan, a canal: Panama"
+Output: true
+Example 2:
+
+Input: "race a car"
+Output: false
+
+```
+
+## 思路
+
+这是一道考察回文的题目，而且是最简单的形式，即判断一个字符串是否是回文。
+
+针对这个问题，我们可以使用头尾双指针，
+
+- 如果两个指针的元素不相同，则直接返回false,
+- 如果两个指针的元素相同，我们同时更新头尾指针，循环。 直到头尾指针相遇。
+
+时间复杂度为O(n).
+
+拿“noon”这样一个回文串来说，我们的判断过程是这样的：
+
+![125.valid-palindrome-1](../assets/problems/125.valid-palindrome-1.png)
+
+拿“abaa”这样一个不是回文的字符串来说，我们的判断过程是这样的：
+
+![125.valid-palindrome-2](../assets/problems/125.valid-palindrome-2.png)
+
+
+
+## 关键点解析
+
+- 双指针
+
+## 代码
+
+```js
+
+/*
+ * @lc app=leetcode id=125 lang=javascript
+ *
+ * [125] Valid Palindrome
+ */
+// 只处理英文字符（题目忽略大小写，我们前面全部转化成了小写， 因此这里我们只判断小写）和数字
+function isValid(c) {
+  const charCode = c.charCodeAt(0);
+  const isDigit =
+    charCode >= "0".charCodeAt(0) && charCode <= "9".charCodeAt(0);
+  const isChar = charCode >= "a".charCodeAt(0) && charCode <= "z".charCodeAt(0);
+
+  return isDigit || isChar;
+}
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+var isPalindrome = function(s) {
+  s = s.toLowerCase();
+  let left = 0;
+  let right = s.length - 1;
+
+  while (left < right) {
+    if (!isValid(s[left])) {
+      left++;
+      continue;
+    }
+    if (!isValid(s[right])) {
+      right--;
+      continue;
+    }
+
+    if (s[left] === s[right]) {
+      left++;
+      right--;
+    } else {
+      break;
+    }
+  }
+
+  return right <= left;
+};
+```

problems/131.palindrome-partitioning.md

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+
+## 题目地址
+
+https://leetcode.com/problems/palindrome-partitioning/description/
+
+## 题目描述
+
+```
+Given a string s, partition s such that every substring of the partition is a palindrome.
+
+Return all possible palindrome partitioning of s.
+
+Example:
+
+Input: "aab"
+Output:
+[
+  ["aa","b"],
+  ["a","a","b"]
+]
+
+```
+
+## 思路
+
+这是一道求解所有可能性的题目， 这时候可以考虑使用回溯法。 回溯法解题的模板我们已经在很多题目中用过了，
+这里就不多说了。大家可以结合其他几道题目加深一下理解。
+
+
+## 关键点解析
+
+- 回溯法
+
+## 代码
+
+```js
+
+
+/*
+ * @lc app=leetcode id=131 lang=javascript
+ *
+ * [131] Palindrome Partitioning
+ */
+
+function isPalindrom(s) {
+    let left = 0;
+    let right = s.length - 1;
+
+    while(left < right && s[left] === s[right]) {
+        left++;
+        right--;
+    }
+
+    return left >= right;
+}
+ function backtrack(s, list, tempList, start) {
+    const sliced = s.slice(start);
+
+    if (isPalindrom(sliced) && (tempList.join("").length === s.length)) list.push([...tempList]);
+
+    for(let i = 0; i < sliced.length; i++) {
+        const sub = sliced.slice(0, i + 1);
+        if (isPalindrom(sub)) {
+            tempList.push(sub);
+        } else {
+            continue;
+        }
+        backtrack(s, list, tempList, start + i + 1);
+        tempList.pop();
+    }
+ }
+/**
+ * @param {string} s
+ * @return {string[][]}
+ */
+var partition = function(s) {
+    // "aab"
+    // ["aa", "b"]
+    // ["a", "a", "b"]
+    const list = [];
+    backtrack(s, list, [], 0);
+    return list;
+};
+
+
+```
+
+## 相关题目
+- [39.combination-sum](./39.combination-sum.md)
+- [40.combination-sum-ii](./40.combination-sum-ii.md)
+- [46.permutations](./46.permutations.md)
+- [47.permutations-ii](./47.permutations-ii.md)
+- [78.subsets](./78.subsets.md)
+- [90.subsets-ii](./90.subsets-ii.md)
+

problems/39.combination-sum.md

@@ -134,3 +134,5 @@ var combinationSum = function(candidates, target) {
 - [47.permutations-ii](./47.permutations-ii.md)
 - [78.subsets](./78.subsets.md)
 - [90.subsets-ii](./90.subsets-ii.md)
+- [131.palindrome-partitioning](./131.palindrome-partitioning.md)
+

problems/40.combination-sum-ii.md

@@ -138,3 +138,4 @@ var combinationSum2 = function(candidates, target) {
 - [47.permutations-ii](./47.permutations-ii.md)
 - [78.subsets](./78.subsets.md)
 - [90.subsets-ii](./90.subsets-ii.md)
+- [131.palindrome-partitioning](./131.palindrome-partitioning.md)

problems/46.permutations.md

@@ -102,4 +102,5 @@ var permute = function(nums) {
 - [47.permutations-ii](./47.permutations-ii.md)
 - [78.subsets](./78.subsets.md)
 - [90.subsets-ii](./90.subsets-ii.md)
+- [131.palindrome-partitioning](./131.palindrome-partitioning.md)
 

problems/47.permutations-ii.md

@@ -104,3 +104,4 @@ var permuteUnique = function(nums) {
 - [46.permutations](./46.permutations.md)
 - [78.subsets](./78.subsets.md)
 - [90.subsets-ii](./90.subsets-ii.md)
+- [131.palindrome-partitioning](./131.palindrome-partitioning.md)

problems/78.subsets.md

@@ -112,5 +112,6 @@ var subsets = function(nums) {
 - [46.permutations](./46.permutations.md)
 - [47.permutations-ii](./47.permutations-ii.md)
 - [90.subsets-ii](./90.subsets-ii.md)
+- [131.palindrome-partitioning](./131.palindrome-partitioning.md)
 
 

problems/90.subsets-ii.md

@@ -112,6 +112,7 @@ var subsetsWithDup = function(nums) {
 - [46.permutations](./46.permutations.md)
 - [47.permutations-ii](./47.permutations-ii.md)
 - [78.subsets](./78.subsets.md)
+- [131.palindrome-partitioning](./131.palindrome-partitioning.md)