[leetcode]Search in Rotated Sorted Array

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.

二分查找的变形

算法思路：

a[mid] == target return mid;

a[mid] < target 分为两种情况

1. a[mid]和target在同半边，begin = mid + 1

2. a[mid]和target在不同的半边，则a[mid]肯定在后面，target在前半边，因此往前找end = mid - 1;

a[mid] > target同理

1. a[mid]和target在同半边，end = mid - 1;

2. a[mid]和target在不同的半边，则a[mid]肯定在前面，target在后半边，因此往前找begin = mid + 1;

代码如下：

public class Solution {
    public int search(int[] a, int target) {
        if(a == null || a.length == 0) return -1;
        int length = a.length;
        int begin = 0, end = length - 1;
        while(begin <= end){
            int mid = (begin + end) >> 1;
            if(a[mid] == target){
                return mid;
            }else if(a[mid] > target){
                if(target < a[begin] && a[mid] >= a[begin]){//这里如果写成a[mid]<=a[end]就不行了，因为a[mid]<=a[end]不能保证在前半段
                    begin = mid + 1;
                }else{
                    end = mid - 1;
                }
            }else{
                if(a[mid] < a[begin] && target >= a[begin]){
                    end = mid - 1;
                }else{
                    begin = mid + 1;
                }
            }
        }
        return -1;
    }
}