feat: azl397985856#33

Author: luzhipeng

README.en.md

@@ -132,6 +132,7 @@ The data structures mainly includes:
 - [0019. Remove Nth Node From End of List](./problems/19.removeNthNodeFromEndofList.md)
 - [0024. Swap Nodes In Pairs](./problems/24.swapNodesInPairs.md)
 - [0029.divide-two-integers](./problems/29.divide-two-integers.md) 🆕
+- [0033.search-in-rotated-sorted-array](./problems/33.search-in-rotated-sorted-array.md) 🆕
 - [0039.combination-sum](./problems/39.combination-sum.md)
 - [0040.combination-sum-ii](./problems/40.combination-sum-ii.md)
 - [0046.permutations](./problems/46.permutations.md)

README.md

@@ -132,6 +132,7 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 - [0019. Remove Nth Node From End of List](./problems/19.removeNthNodeFromEndofList.md)
 - [0024. Swap Nodes In Pairs](./problems/24.swapNodesInPairs.md)
 - [0029.divide-two-integers](./problems/29.divide-two-integers.md) 🆕
+- [0033.search-in-rotated-sorted-array](./problems/33.search-in-rotated-sorted-array.md) 🆕
 - [0039.combination-sum](./problems/39.combination-sum.md)
 - [0040.combination-sum-ii](./problems/40.combination-sum-ii.md)
 - [0046.permutations](./problems/46.permutations.md)

assets/drawio/33.search-in-rotated-sorted-array.drawio

@@ -0,0 +1 @@
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assets/problems/search-in-rotated-sorted-array-1.jpg

@@ -0,0 +1,122 @@
+## 题目地址
+https://leetcode.com/problems/search-in-rotated-sorted-array/
+
+## 题目描述
+
+```
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
+
+You are given a target value to search. If found in the array return its index, otherwise return -1.
+
+You may assume no duplicate exists in the array.
+
+Your algorithm's runtime complexity must be in the order of O(log n).
+
+Example 1:
+
+Input: nums = [4,5,6,7,0,1,2], target = 0
+Output: 4
+Example 2:
+
+Input: nums = [4,5,6,7,0,1,2], target = 3
+Output: -1
+
+```
+
+## 思路
+
+这是一个我在网上看到的前端头条技术终面的一个算法题。
+
+题目要求时间复杂度为logn，因此基本就是二分法了。 这道题目不是直接的有序数组，不然就是easy了。
+
+首先要知道，我们随便选择一个点，将数组分为前后两部分，其中一部分一定是有序的。
+
+具体步骤：
+
+- 我们可以先找出mid，然后根据mid来判断，mid是在有序的部分还是无序的部分
+
+假如mid小于start，则mid一定在右边有序部分。
+假如mid大于等于start， 则mid一定在左边有序部分。
+
+> 注意等号的考虑
+
+- 然后我们继续判断target在哪一部分， 我们就可以舍弃另一部分了
+
+我们只需要比较target和有序部分的边界关系就行了。 比如mid在右侧有序部分，即[mid, end]
+那么我们只需要判断 target >= mid && target <= end 就能知道target在右侧有序部分，我们就
+可以舍弃左边部分了(start = mid + 1)， 反之亦然。
+
+我们以([6,7,8,1,2,3,4,5], 4)为例讲解一下：
+
+![search-in-rotated-sorted-array-1](../assets/problems/search-in-rotated-sorted-array-1.jpg)
+
+
+![search-in-rotated-sorted-array-1](../assets/problems/search-in-rotated-sorted-array-2.jpg)
+
+
+## 关键点解析
+
+- 二分法
+- 找出有序区间，然后根据target是否在有序区间舍弃一半元素
+## 代码
+
+* 语言支持: Javascript
+
+```js
+/*
+ * @lc app=leetcode id=33 lang=javascript
+ *
+ * [33] Search in Rotated Sorted Array
+ */
+/**
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number}
+ */
+var search = function(nums, target) {
+  // 时间复杂度：O(logn)
+  // 空间复杂度：O(1)
+  // [6,7,8,1,2,3,4,5]
+  let start = 0;
+  let end = nums.length - 1;
+
+  while (start <= end) {
+    const mid = start + ((end - start) >> 1);
+    if (nums[mid] === target) return mid;
+
+    // [start, mid]有序
+
+    // ️⚠️注意这里的等号
+    if (nums[mid] >= nums[start]) {
+      //target 在 [start, mid] 之间
+
+      // 其实target不可能等于nums[mid]， 但是为了对称，我还是加上了等号
+      if (target >= nums[start] && target <= nums[mid]) {
+        end = mid - 1;
+      } else {
+        //target 不在 [start, mid] 之间
+        start = mid + 1;
+      }
+    } else {
+      // [mid, end]有序
+
+      // target 在 [mid, end] 之间
+      if (target >= nums[mid] && target <= nums[end]) {
+        start = mid + 1;
+      } else {
+        // target 不在 [mid, end] 之间
+        end = mid - 1;
+      }
+    }
+  }
+
+  return -1;
+};
+```
+
+## 扩展
+
+
+