## [Leetcode]33. Search in Rotated Sorted Array(C++)

### 题目描述

There is an integer array `nums` sorted in ascending order (with distinct values).

Prior to being passed to your function, `nums` is rotated at an unknown pivot index `k (0 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (0-indexed). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index `3` and become `[4,5,6,7,0,1,2]`.

Given the array `nums` after the rotation and an integer `target`, return the index of `target` if it is in `nums`, or `-1` if it is not in `nums`.

### 例子

#### 例子 1

Input: `nums = [4,5,6,7,0,1,2], target = 0` Output: `4`

#### 例子 2

Input: `nums = [4,5,6,7,0,1,2], target = 3` Output: `-1`

#### 例子 3

Input: `nums = [1], target = 0` Output: `-1`

### Constraints

• `1 <= nums.length <= 5000`
• `-10^4 <= nums[i] <= 10^4`
• All values of `nums` are unique.
• `nums` is guaranteed to be rotated at some pivot.
• `-10^4 <= target <= 10^4`

### 解题思路

``````class Solution {
public:
int search(vector<int>& nums, int target) {
int left = 0;
int right = nums.size() - 1;
while (left <= right) {
int mid = (left + right) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] < target) {
if (nums[right] < target && nums[mid] <= nums[right]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else {
if (nums[left] > target && nums[mid] >= nums[left]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
}

return -1;
}
};
``````
• 时间复杂度: `O(log n)`
• 空间复杂度: `O(1)`

GitHub 代码同步地址： 33.SearchInRotatedSortedArray.cpp

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