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;
    }
};