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