33. Search in Rotated Sorted Array-优快云博客

本文介绍了一种在未知旋转点的有序数组中查找目标值的方法。文章提供了两种使用二分查找法解决问题的实现方案，并详细解释了每一步的判断条件。

Search in Rotated Sorted Array

题意：

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.
主要分下面三种情况，看target在那那个位置：
这里写图片描述

其实这道题直接用for循环遍历一遍也可以，但考虑到使用二分法，则用如下：

代码1：

public class Solution {  
    public int search(int[] A, int target) {  
        int l = 0;  
        int r = A.length - 1;  
        while (l <= r) {  
            int mid = (l + r) / 2;  
            if (target == A[mid]) return mid;  
            if (A[l] <= A[r]) {  //one
                if (target < A[mid]) r = mid - 1;  
                else l = mid + 1;  
            } else if (A[l] <= A[mid]) { //two 
                if (target > A[mid] || target < A[l]) l = mid + 1;  
                else r = mid - 1;  
            } else {  //three
                if (target < A[mid] || target > A[r]) r = mid - 1;  
                else l = mid + 1;  
            }  
        }  
        return -1;  
    }  
}

代码2：

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

            if (nums[start] <= nums[mid]){
                 if (target < nums[mid] && target >= nums[start]) 
                    end = mid - 1;
                 else
                    start = mid + 1;
            } 

            if (nums[mid] <= nums[end]){
                if (target > nums[mid] && target <= nums[end])
                    start = mid + 1;
                 else
                    end = mid - 1;
            }
        }
        return -1;
    }
}