## Search in Rotated Sorted Array II

### 描述

Follow up for "Search in Rotated Sorted Array": What if **duplicates** are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

### 分析

* 若`A[left] < A[mid]`，则区间`[left,mid]`一定递增
* 若`A[left] == A[mid]` 确定不了，那就`left++`，往下看一步即可。

### 代码

{% if book.java %}
```java
// Search in Rotated Sorted Array II
// Time Complexity: O(n)，Space Complexity: O(1)
public class Solution {
public boolean search(int[] nums, int target) {
int first = 0, last = nums.length;
while (first != last) {
final int mid = first + (last - first) / 2;
if (nums[mid] == target)
return true;
if (nums[first] < nums[mid]) {
if (nums[first] <= target && target < nums[mid])
last = mid;
else
first = mid + 1;
} else if (nums[first] > nums[mid]) {
if (nums[mid] < target && target <= nums[last-1])
first = mid + 1;
else
last = mid;
} else
//skip duplicate one
first++;
}
return false;
}
};
```
{% endif %}

{% if book.cpp %}
```cpp
// Search in Rotated Sorted Array II
// Time Complexity: O(n)，Space Complexity: O(1)
class Solution {
public:
bool search(const vector& nums, int target) {
int first = 0, last = nums.size();
while (first != last) {
const int mid = first + (last - first) / 2;
if (nums[mid] == target)
return true;
if (nums[first] < nums[mid]) {
if (nums[first] <= target && target < nums[mid])
last = mid;
else
first = mid + 1;
} else if (nums[first] > nums[mid]) {
if (nums[mid] < target && target <= nums[last-1])
first = mid + 1;
else
last = mid;
} else
//skip duplicate one
first++;
}
return false;
}
};
```
{% endif %}

### 相关题目

* [Search in Rotated Sorted Array](search-in-rotated-sorted-array.md)
* [Find Minimum in Rotated Sorted Array](find-minimum-in-rotated-sorted-array.md)
* [Find Minimum in Rotated Sorted Array II](find-minimum-in-rotated-sorted-array-ii.md)