[Leetcode] Search in Rotated Sorted Array

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.

Solution:

public class Solution {
    public int search(int[] A, int target) {
        if(A.length<=0)
            return 0;
        int start=0;
        int end=A.length-1;
        while(start<=end){
            int mid=(start+end)/2;
            if(A[mid]==target)
                return mid;
            if(A[mid]>=A[start]){
                if(A[mid]>=target&&target>=A[start]){
                    end=mid-1;
                }
                else
                    start=mid+1;
            }else if(A[mid]<=A[end]){
                if(A[mid]<=target&&target<=A[end]){
                    start=mid+1;
                }else
                    end=mid-1;
            }
        }
        return -1;
    }
}

Note:

黄色部分，不要少加了=号，这样在某些情况下（比如[1], 0的情况）会进入不了判断条件里，导致start和end无法改变，超时。