leetcode Search in Rotated Sorted Array

class Solution {public:    int search(vector<int>& nums, int target) {        if(nums.size()==1){if(target==nums[0]) return 0;else return -1;}        if(nums.size()==2){if(target==nums[0]) return 0;else if(target==nums[1]) return 1;else return -1;}        int changed=0;        int pivot=0;        for(int i=2;i<nums.size();i++){            if((nums[i]-nums[i-1])*(nums[i-1]-nums[i-2])<0){                changed=1;pivot=i;break;            }        }        int l=nums.size();        if(changed==0){return binarySearch(nums,target,0,l-1);}        if(binarySearch(nums,target,0,pivot-1)!=-1){return binarySearch(nums,target,0,pivot-1);}        else return binarySearch(nums,target,pivot,l-1);    }private:    int binarySearch(vector<int> &nums,int target,int a,int b){        if(nums.size()==0) return -1;        if(a==b){            if(target==nums[a]){return a;}            else{return -1;}        }if(b-a==1){if(target==nums[a]){return a;}else if(target==nums[b]){return b;}else return -1;}        if(target==nums[(a+b)/2]){            return (a+b)/2;        }        if(target>nums[(a+b)/2]){            return binarySearch(nums,target,((a+b)/2)+1,b);        }        if(target<nums[(a+b)/2]){            return binarySearch(nums,target,a,((a+b)/2)-1);        }    }};

查找可采用二分，时间复杂度为logN，然而二分的前提为数组是有序的，此时需找出有序数列的断点，断点的左右序列都是有序的，分别进行二分查找即可。