[LeetCode] 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.

You may assume no duplicate exists in the array.

• 解题思路
  二分查找
• 实现代码

/*************************************************************
    *  @Author   : 楚兴
    *  @Date     : 2015/2/8 17:17
    *  @Status   : Accepted
    *  @Runtime  : 10 ms
*************************************************************/
#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
    int findMin(vector<int> &num) {
        int begin = 0;
        int end = num.size() - 1;
        int mid;
        while (begin + 1 < end)
        {
            mid = (begin + end) / 2;
            if (num[mid] > num[end])
            {
                begin = mid;
            }
            else
            {
                end = mid;
            }
        }
        return num[begin] < num[end] ? num[begin] : num[end];
    }
};

int main()
{
    Solution s;
    vector<int> nums;
    nums.push_back(4);
    nums.push_back(5);
    nums.push_back(6);
    nums.push_back(7);
    nums.push_back(1);
    nums.push_back(2);
    nums.push_back(3);
    cout<<s.findMin(nums)<<endl;
    system("pause");
}

• 根据官方解答优化后的代码：

/*************************************************************
    *  @Author   : 楚兴
    *  @Date     : 2015/2/8 17:46
    *  @Status   : Accepted
    *  @Runtime  : 7 ms
*************************************************************/
#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
    int findMin(vector<int> &num) {
        int begin = 0;
        int end = num.size() - 1;
        int mid;
        while (begin < end)
        {
            mid = (begin + end) / 2;
            if (num[mid] > num[end])
            {
                begin = mid + 1;
            }
            else
            {
                end = mid;
            }
        }
        return num[begin];
    }
};