[leetcode] 33. Search in Rotated Sorted Array

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 

0 1 2 4 5 6 7

 might become 

4
5 6 7 0 1 2

).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

解法一：

二分法，先确定那边是ordered数组。

class Solution {
public:
int 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 mid;
else if(nums[mid]<nums[right]){
if(target>nums[mid] && target<=nums[right]) left = mid +1;
else right = mid -1;

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

}
return -1;

}
};