Leetcode 33. Search in Rotated Sorted Array

题目如下：

Suppose an array sorted in ascending order 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.

即对于一个原始的非递减非重复序列，在某处旋转后变成两段有序区，左边区域值较大，右边区域值较小，给定一个target求其位置。

对于有序序列问题一般就是二分查找。

设区间a[low,high]，其中间位置mid=(low+high)/2,下面分情况讨论：

1 target==a[mid]返回mid即可;

2 when a[low]<a[high]，说明该区间已经是有序区，按照常见的有序区二分即可：

    if target>a[mid],low=mid+1;

    if target<a[mid],high=mid-1;

3 when a[mid]>=a[low]，由旋主序列特性可知，此时中间位置在左边较大的序列

    if target<a[low]，说明target比左序区a[low...mid]值都小,low=mid+1;

    if target>a[mid]，说明target比a[loq...mid]都要大，low=mid+1;

    else a[low]<=target<a[mid]，high=mid-1;

4 when a[mid]<a[low]，由旋主序列特性可知，此时中间位置在右边较小的序列

    if target>a[high]，说明target比右序区a[mid...high]值都大,high=mid-1;

    if target<a[mid]，说明target比a[mid...high]都要小，high=mid-1;

    else a[mid]<target<=a[high]，low=mid+1;

int search(vector<int>& nums,int target){    int low=0,high=nums.size()-1;    int mid;    while(low<=high){mid=(low+high)/2;if(target==nums[mid])return mid;if(nums[low]<nums[high]){if(target>nums[mid])low=mid+1;                         if(target<nums[mid])high=mid-1;                         }//low-high之间都是有序else{    if(nums[mid]>=nums[low]){if(target>nums[mid]||target<nums[low]) low=mid+1;else high=mid-1;                          }//mid处于左侧较大的有序区    else{if(target<nums[mid]||target>nums[high])high=mid-1;else low=mid+1;         }//mid处于右侧较小区域    }//Low-high之间是由两段有序区组成                      }    if(low==high&&nums[low]==target)return low;    return -1;