## [Leetcode]81. Search in Rotated Sorted Array II(C++)

### 题目描述

There is an integer array `nums` sorted in non-decreasing order (not necessarily with distinct values).

Before 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, `[0,1,2,4,4,4,5,6,6,7]` might be rotated at pivot index `5` and become `[4,5,6,6,7,0,1,2,4,4]`.

Given the array `nums` after the rotation and an integer `target`, return `true` if `target` is in `nums`, or `false` if it is not in `nums`.

### 例子

#### 例子 1

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

#### 例子 2

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

This problem is the same as `Search in Rotated Sorted Array`, where `nums` may contain duplicates. Would this affect the runtime complexity? How and why?

### Constraints

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

### 解题思路

• `Left = Left + 1 = ... = Mid ( = ... = Right)`，此时左侧区间元素全部相等（并且有可能和右侧末尾部分元素相等），`Target` 在右侧区间
• `Mid = Mid + 1 = ... = Right = Left`，此时右侧区间元素全部相等，并且和左侧前面部分元素相等， `Target` 在左侧区间

`Target > Mid` 时思路类似，不再赘述。

``````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 true;
} else if (nums[mid] < target) {
if (nums[mid] == nums[right]) {
right--;
} else if (nums[right] < target && nums[mid] < nums[right]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else {
if (nums[mid] == nums[left]) {
left++;
} else if (nums[left] > target && nums[mid] > nums[left]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
}

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

GitHub 代码同步地址： 81.SearchInRotatedSortedArrayIi.cpp

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