leetcode之Search in Rotated Sorted Array

题目：

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.

解答：

利用二分搜索，直接看代码就懂了

class Solution {public:    int search(vector<int>& nums, int target) {        int l = 0;        int r = nums.size() - 1;        while(l <= r)        {            int mid = (l + r) >> 1;            if(nums[mid] == target)                return mid;            if(nums[l] <= nums[mid])//如果前半段是顺序的            {                if(target <= nums[mid] && target >= nums[l])//此时target在前半段                    r = mid - 1;                else                    l = mid + 1;            }            else if(nums[mid] <= nums[r])//后半段是顺序的            {                if(target <= nums[r] && target >= nums[mid])//次数target在后半段，注意要有等于号                    l = mid + 1;                else                    r = mid - 1;            }        }        return -1;    }};