[81]Search in Rotated Sorted Array II

【题目描述】

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.
【思路】

和Search in Rotated Sorted Array相比，主要是要考虑有重复的情况，还是分成三种情况考虑：

1.nums[low]<nums[mid],mid在左半部分

2.nums[low]>nums[mid]，mid在右半部分

3..nums[low]==nums[mid]，有重复 

【代码】

class Solution {
public:
bool search(vector<int>& nums, int target) {
int n=nums.size();
int low,high,mid;
low=0;
high=n;
while(low<high){
mid=(low+high)/2;
if(nums[mid]==target) return true;
if(nums[low]<nums[mid]){
if(nums[low]<=target&&nums[mid]>target) high=mid;
else low=mid+1;
}
else if(nums[low]>nums[mid]){
if(nums[mid]<target&&nums[high-1]>=target) low=mid+1;
else high=mid;
}
else{
low++;
}
}
return false;
}
};