[leetcode] 81. Search in Rotated Sorted Array II

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.

解法一：

当nuts[mid] == nums[right]的时候，right向左移动一位。其他的和Search in Rotated Sorted Array问题中没有duplicate的代码类似。

class Solution {
public:
bool search(vector<int>& nums, int target) {
if(nums.empty()) return 0;
int left = 0, right = nums.size()-1;

while(left<=right){
int mid = (left+right)/2;
if(target== nums[mid]) return true;
else if(nums[mid]<nums[right]){
if(target>nums[mid] && target<=nums[right]) left = mid +1;
else right = mid -1;
}
else if (nums[mid]>nums[right]){
if(target>=nums[left] && target < nums[mid]) right = mid -1;
else left = mid + 1;
} else --right;
}
return false;
}
};