leetcode-- Container With Most Water

最新推荐文章于 2022-03-29 21:43:55 发布

nianhuatingyu

最新推荐文章于 2022-03-29 21:43:55 发布

阅读量220

点赞数

CC 4.0 BY-SA版权

分类专栏：面试题 leetcode 文章标签： leetcode 面试题

本文链接：https://blog.youkuaiyun.com/nianhuatingyu/article/details/52484887

面试题同时被 2 个专栏收录

17 篇文章

订阅专栏

leetcode

6 篇文章

订阅专栏

本文介绍了一种寻找能容纳最多水的两个垂直线的算法。该算法通过双指针技术，从两端向中间搜索，逐步缩小搜索范围，直到找到最优解。文章提供了完整的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container.

Subscribe to see which companies asked this question

/**********************************************
*@time 2016/09/09 10:36
*@place Ctrip.15#.9f
*@author  dzq
**********************************************/
#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
class Solution
{
public:
    int maxArea(vector<int>& height)
    {
       int lift=0;
       int right=height.size()-1;


       int cap=min(height[lift],height[right])*(right-lift);
       while(lift<right)
       {
           int water=min(height[lift],height[right])*(right-lift);
           if(cap<water) cap=water;
           //比原来容器更大的情况可能只有一种:
           //在lift和right中有一个比=min(height[lift],height[right])更大的高度
           if(height[lift]<height[right])
           {
               lift++;
               continue;
           }
           else
           {
               right--;
               continue;
           }
       }
       cout<<cap<<endl;
       return cap;
    }
};
int main()
{
    Solution solution;
    vector<int> nums;
    nums.push_back(1);
    nums.push_back(1);
    solution.maxArea(nums);
    return 0;
}