LeetCode 33 Search in Rotated Sorted Array

原题:（频率3）

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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.

题意：一个有序数组从某个地方旋转，给定一个数字，找到返回位置，否则返回-1

代码和思路：

//简单二分查找的应用
class Solution {
    public int search(int[] nums, int target) {
        
        int lo = 0 , hi = nums.length - 1;
        
        while(lo<=hi){
            int mid = (lo+hi)/2;
            if(nums[mid] == target) return mid;
            
            //左边升序
            if(nums[mid] >= nums[lo]){
                if(target >= nums[lo] && target  < nums[mid]){
                    hi = mid - 1;
                }
                else{
                    lo = mid + 1;
                }
            }
            //右边升序
            else{
                if(target > nums[mid] && nums[hi] >= target){
                    lo = mid +1;
                }
                else {
                    hi = mid - 1;
                }
            }
            
        }
        return -1;            
    }
}