Search in Rotated Sorted Array -- Leetcode

Suppose a sorted array 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.

class Solution {public:    int search(int A[], int n, int target) {        int hi,low;        low=0;        hi=n-1;        while(low<=hi){            int mid=(hi+low)/2;            if(A[mid]==target)                return mid;            else if(A[low]<=A[mid]){                if(A[low]<=target && target<A[mid])                    hi=mid-1;                else                     low=mid+1;            }            else{                if(A[mid]<target && target<=A[hi])                    low=mid+1;                else                    hi=mid-1;            }        }        return -1;    }};

总结：

1. 关键思路：二分搜索

2. 先判断前半部分是不是递增的，当是递增时，把搜索区间变为前半部分，看target是否在区间内，如果不在怎把搜索区间设为后半部分。

3. 当前半部分不是递增的，判断target是否在后半区间内，若在在搜索区间设为该区间，否则在另外区间继续搜索。