fix: 文件名重命名

Author: lucifer

README.md

@@ -142,9 +142,9 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 
 #### 简单难度
 
-- [0001.TwoSum](./problems/1.TwoSum.md)
-- [0020.Valid Parentheses](./problems/20.validParentheses.md)
-- [0021.MergeTwoSortedLists](./problems/21.MergeTwoSortedLists.md)
+- [0001.two-sum](./problems/1.two-sum.md)
+- [0020.Valid Parentheses](./problems/20.valid-parentheses.md)
+- [0021.MergeTwoSortedLists](./problems/21.merge-two-sorted-lists.md)
 - [0026.remove-duplicates-from-sorted-array](./problems/26.remove-duplicates-from-sorted-array.md)
 - [0053.maximum-sum-subarray](./problems/53.maximum-sum-subarray-cn.md)
 - [0088.merge-sorted-array](./problems/88.merge-sorted-array.md)
@@ -181,14 +181,14 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 
 #### 中等难度
 
-- [0002. Add Two Numbers](./problems/2.addTwoNumbers.md)
-- [0003. Longest Substring Without Repeating Characters](./problems/3.longestSubstringWithoutRepeatingCharacters.md)
+- [0002.add-two-numbers](./problems/2.add-two-numbers.md)
+- [0003.longest-substring-without-repeating-characters](./problems/3.longest-substring-without-repeating-characters.md)
 - [0005.longest-palindromic-substring](./problems/5.longest-palindromic-substring.md)
 - [0011.container-with-most-water](./problems/11.container-with-most-water.md)
 - [0015.3-sum](./problems/15.3-sum.md)
 - [0017.Letter-Combinations-of-a-Phone-Number](./problems/17.Letter-Combinations-of-a-Phone-Number.md) 🆕
 - [0019. Remove Nth Node From End of List](./problems/19.removeNthNodeFromEndofList.md)
-- [0022.GenerateParentheses](./problems/22.GenerateParentheses.md) 🆕
+- [0022.generate-parentheses.md](./problems/22.generate-parentheses.md) 🆕
 - [0024. Swap Nodes In Pairs](./problems/24.swapNodesInPairs.md)
 - [0029.divide-two-integers](./problems/29.divide-two-integers.md)
 - [0031.next-permutation](./problems/31.next-permutation.md)

problems/1.TwoSum.en.md problems/1.two-sum.en.md

@@ -1,7 +1,9 @@
 ## Problem
+
 https://leetcode-cn.com/problems/two-sum
 
 ## Problem Description
+
 ```
 Given an array of integers, return indices of the two numbers such that they add up to a specific target.
 
@@ -16,15 +18,18 @@ return [0, 1].
 ```
 
 ## Solution
+
 The easiest solution to come up with is Brute Force. We could write two for-loops to traverse every element, and find the target numbers that meet the requirement. However, the time complexity of this solution is O(N^2), while the space complexity is O(1). Apparently, we need to find a way to optimize this solution since the time complexity is too high. What we could do is to record the numbers we have traversed and the relevant index with a Map. Whenever we meet a new number during traversal, we go back to the Map and check whether the `diff` between this number and the target number appeared before. If it did, the problem has been solved and there's no need to continue.
 
 ## Key Points
- - Find the difference instead of the sum
- - Connect every number with its index through the help of Map
- - Less time by more space. Reduce the time complexity from O(N) to O(1)
 
- ## Code
-  - Support Language: JS
+- Find the difference instead of the sum
+- Connect every number with its index through the help of Map
+- Less time by more space. Reduce the time complexity from O(N) to O(1)
+
+## Code
+
+- Support Language: JS
 
 ```js
 /**
@@ -33,18 +38,18 @@ The easiest solution to come up with is Brute Force. We could write two for-loop
  * @return {number[]}
  */
 const twoSum = function (nums, target) {
-    const map = new Map();
-    for (let i = 0; i < nums.length; i++) {
-        const diff = target - nums[i];
-        if (map.has(diff)) {
-            return [map.get(diff), i];
-        }
-        map.set(nums[i], i);
+  const map = new Map();
+  for (let i = 0; i < nums.length; i++) {
+    const diff = target - nums[i];
+    if (map.has(diff)) {
+      return [map.get(diff), i];
     }
-}
+    map.set(nums[i], i);
+  }
+};
 ```
 
-***Complexity Anlysis***
+**_Complexity Anlysis_**
 
-- *Time Complexity*: O(N)
-- *Space Complexity*：O(N)
+- _Time Complexity_: O(N)
+- _Space Complexity_：O(N)