Search in Rotated Sorted Array

33. Search in Rotated Sorted Array

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.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

思路：二分查找，通过中间mid来确定前面是有序的还是后面有序的：如果mid比high高，说明前面是有序的；如果mid比high低，说明后面有序的。在有序的里面作判断，来重新给出low和high的位置。

class Solution {
public:
    int search(vector& nums, int target) {
        int low=0;
        int high=nums.size()-1;
        if(nums.size()==0)
            return -1;
        while(low<=high)
        {
            int mid=(low+high)/2;
            if(nums[mid]==target)
                return mid;
            else if(nums[mid]nums[high])
            {
                //前面有序的
                if(nums[mid]>target&&target>=nums[low])
                {
                    high=mid-1;
                }
                else 
                    low=mid+1;
                
            }
            else 
                return -1;
            
            
        }
        return -1;
        
    }
};