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】

分状况讨论，数组可能有如下三种状况：

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

而后，再看每一种状况中，target在左边仍是在右边，其中第一种状况还能够直接判断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;
}

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