Search in Rotated Sorted Array

查找旋转排序数组

题目

Suppose an array sorted in ascending order 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.

Your algorithm’s runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

解析重点

1.依题我们可以知道，当我们使用二分法的时候，肯定会有一边是排序数组，如果target在排序数组里面，那么继续二分查找，如果不在，那么循环整个逻辑，继续在
另一边查找。

java代码

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]){//如果左边是排序数组，target在左边，那么继续二分查找，不在的话在右边继续查找。
                 if (target < nums[mid] && target >= nums[start]) 
                    end = mid - 1;
                 else
                    start = mid + 1;
            } 
        
            if (nums[mid] <= nums[end]){//与上面逻辑相同
                if (target > nums[mid] && target <= nums[end])
                    start = mid + 1;
                 else
                    end = mid - 1;
            }
        }
        return -1;
    }
}