LeetCode（153） Find Minimum in Rotated Sorted Array

题目

Total Accepted: 65121 Total Submissions: 190974 Difficulty: Medium
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.

分析

二分搜索的扩展应用。

寻找一个有序序列经旋转后的最小元素。

关键是掌握好，有序序列旋转后的元素之间的关系。

AC代码

class Solution {
public:
    int findMin(vector<int>& nums) {
        if (nums.empty())
            return INT_MIN;

        //元素个数
        int size = nums.size();

        int lhs = 0, rhs = size - 1;
        while (lhs <= rhs)
        {
            //递增状态，返回低位值
            if (nums[lhs] <= nums[rhs])
                return nums[lhs];
            //相邻，返回较小值
            else if (rhs - lhs == 1)
                return nums[lhs] < nums[rhs] ? nums[lhs] : nums[rhs];

            //判断子序列
            int mid = (lhs + rhs) / 2;
            //右侧递增，则最小值位于左半部分
            if (nums[mid] < nums[rhs])
            {
                rhs = mid;
            }
            //否则，最小值位于右半部分
            else{
                lhs = mid + 1;
            }
        }//while
    }

    //优化函数
    int findMin2(vector<int>& nums)
    {
        if (nums.size() == 1)
            return nums[0];

        int lhs = 0, rhs = nums.size() - 1;

        while (nums[lhs] > nums[rhs])
        {
            int mid = (lhs + rhs) / 2;
            if (nums[mid] > nums[rhs])
                lhs = mid + 1;
            else
                rhs = mid;
        }//while
        return nums[lhs];
    }
};

GitHub测试程序源码