Search in Rotated Sorted Array

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

(i.e., 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.

【二分思路】

分情况讨论，数组可能有以下三种情况：

来源网络：http://blog.csdn.net/ljiabin/article/details/40453607

然后，再看每一种情况中，target在左边还是在右边，其中第一种情况还可以直接判断target有可能不在数组范围内。

代码：

int search(int* nums, int numsSize, int target) {
    int low=0;
    int high=numsSize-1;
    if(low>high)return -1;
    while(low<=high){
        int mid=(low+high)/2;
        if(nums[mid]==target)return mid;
        if(nums[low]<=nums[high]){
            if(nums[mid]>target)high=mid-1;
            else low=mid+1;
        }
        else if(nums[low]<=nums[mid]){
            if(target<nums[low]||target>nums[mid])low=mid+1;
            else high=mid-1;
        }
        else{
            if(target<nums[mid]||target>=nums[low])high=mid-1;
            else low=mid+1;
        }
    }
    return -1;
}