lintcode:Find Minimum in Rotated Sorted Array

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).

Find the minimum element.

Example
Given [4, 5, 6, 7, 0, 1, 2] return 0

Note
You may assume no duplicate exists in the array.

class Solution {public:    /**     * @param num: a rotated sorted array     * @return: the minimum number in the array     */    int findMin(vector<int> &num) {        // write your code here        int l=0,r=num.size()-1;        while(l<r-1){//           int mid=l+(r-l)/2;           if(num[mid]<num[r]){               r=mid;           }else{               l=mid;           }        }        if(num[l]<num[r]){            return num[l];        }else{            return num[r];        }    }};

这题说实话比较难。
关键在于要跟num[r]比较，跟num[l]比较比较复杂；
二、在于循环条件l< r-1