feat: 增加题目 301， 518

README.md

@@ -56,11 +56,13 @@ leetcode 题解，记录自己的 leecode 解题之路。
 - [209.minimum-size-subarray-sum](./problems/209.minimum-size-subarray-sum.md)
 - [900.rle-iterator](./problems/900.rle-iterator.md)
 - [322.coin-change](./problems/322.coin-change.md)
+- [518.coin-change-2](./problems/518.coin-change-2.md)
 
 #### 困难难度
 
 - [145.binary-tree-postorder-traversal](./problems/145.binary-tree-postorder-traversal.md)
 - [146.lru-cache](./problems/146.lru-cache.md)
+- [301.remove-invalid-parentheses](./problems/301.remove-invalid-parentheses.md)
 
 ### 数据结构与算法的总结
 
@@ -78,6 +80,4 @@ TODO
 
 [301.remove-invalid-parentheses]
 
-[609.find-duplicate-file-in-system]
-
 

problems/301.remove-invalid-parentheses.md

@@ -0,0 +1,137 @@
+
+## 题目地址
+https://leetcode.com/problems/remove-invalid-parentheses/description/
+
+## 题目描述
+```
+Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.
+
+Note: The input string may contain letters other than the parentheses ( and ).
+
+Example 1:
+
+Input: "()())()"
+Output: ["()()()", "(())()"]
+Example 2:
+
+Input: "(a)())()"
+Output: ["(a)()()", "(a())()"]
+Example 3:
+
+Input: ")("
+Output: [""]
+
+```
+
+## 思路
+
+我们的思路是先写一个函数用来判断给定字符串是否是有效的。 然后再写一个函数，这个函数
+依次删除第i个字符，判断是否有效，有效则添加进最终的返回数组。
+
+这样的话实现的功能就是， 删除`一个` 小括号使之有效的所有可能。因此只需要递归调用`依次删除第i个字符`的功能就可以了。
+
+而且由于题目要求是要删除最少的小括号，因此我们的思路是使用广度优先遍历，而不是深度有限的遍历。
+
+## 关键点解析
+
+- 广度有限遍历
+
+- 使用队列简化操作
+
+- 使用一个visited的mapper， 来避免遍历同样的字符串
+
+
+## 代码
+```js
+/*
+ * @lc app=leetcode id=301 lang=javascript
+ *
+ * [301] Remove Invalid Parentheses
+ *
+ * https://leetcode.com/problems/remove-invalid-parentheses/description/
+ *
+ * algorithms
+ * Hard (38.52%)
+ * Total Accepted:    114.3K
+ * Total Submissions: 295.4K
+ * Testcase Example:  '"()())()"'
+ *
+ * Remove the minimum number of invalid parentheses in order to make the input
+ * string valid. Return all possible results.
+ *
+ * Note: The input string may contain letters other than the parentheses ( and
+ * ).
+ *
+ * Example 1:
+ *
+ *
+ * Input: "()())()"
+ * Output: ["()()()", "(())()"]
+ *
+ *
+ * Example 2:
+ *
+ *
+ * Input: "(a)())()"
+ * Output: ["(a)()()", "(a())()"]
+ *
+ *
+ * Example 3:
+ *
+ *
+ * Input: ")("
+ * Output: [""]
+ *
+ */
+var isValid = function(s) {
+  let openParenthes = 0;
+  for(let i = 0; i < s.length; i++) {
+    if (s[i] === '(') {
+      openParenthes++;
+    } else if (s[i] === ')') {
+      if (openParenthes === 0) return false;
+      openParenthes--;
+    }
+  }
+  return openParenthes === 0;
+};
+/**
+ * @param {string} s
+ * @return {string[]}
+ */
+var removeInvalidParentheses = function(s) {
+  if (!s || s.length === 0) return [""];
+  const ret = [];
+  const queue = [s];
+  const visited = {};
+  let current = null;
+  let removedParentheses = 0; // 只记录最小改动
+
+  while ((current = queue.shift())) {
+    let hit = isValid(current);
+    if (hit) {
+      if (!removedParentheses) {
+       removedParentheses =  s.length - current.length
+      }
+      if (s.length - current.length > removedParentheses) return ret.length === 0 ? [""] : ret;;
+      ret.unshift(current);
+      continue;
+    }
+    for (let i = 0; i < current.length; i++) {
+      if (current[i] !== ')' && current[i] !== '(') continue;
+      const subString = current.slice(0, i).concat(current.slice(i + 1));
+      if (visited[subString]) continue;
+      visited[subString] = true;
+      queue.push(subString);
+    }
+  }
+
+  return ret.length === 0 ? [""] : ret;
+};
+```
+
+## 扩展
+
+相似问题:
+
+[validParentheses](./validParentheses.md)

problems/322.coin-change.md

@@ -129,3 +129,7 @@ var coinChange = function(coins, amount) {
 ## 扩展
 
 这是一道很简单描述的题目， 因此很多时候会被用到大公司的电面中。
+
+相似问题:
+
+[518.coin-change-2](./518.coin-change-2.md)