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;    }};