LeetCode 33 Search in Rotated Sorted Array（在旋转排序数组中搜索）（*）

翻译

假定一个数组在一个我们预先不知道的轴点旋转。

例如，0 1 2 4 5 6 7可能会变为4 5 6 7 0 1 2。

给你一个目标值去搜索，如果找到了则返回它的索引，否则返回-1。

你可以假定没有重复的元素存在于数组中。

原文

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.

代码

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int l = 0, r = nums.size()-1;
        while (l<=r) {
            int mid = (r-l)/2+l;
            if (nums[mid] == target)
                return mid;
            if (nums[mid] < nums[r]) {
                if (nums[mid]<target && target<=nums[r])
                    l = mid+1;
                else
                    r = mid-1;
            } else {
                if(nums[l]<=target && target<nums[mid])
                    r = mid-1;
                else
                    l = mid+1;
            }
        }
        return -1;
    }
};