【LeetCode】 11. Container With Most Water 【Python || C++】双语言实现

题目（来自leetcode网站）：

这个题目的意思是，给你排好一堆柱子，找出容量最大的两根；

最基本的想法是用两个for循环，但是这样太慢了；一开始没想出来，后来看了别人的想法。。。

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 and n is at least 2.

 

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

 

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

#Python版本
class Solution:
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        container=0
        begin_h,end_h=0,len(height)-1
        while begin_h<end_h:
            container = (end_h-begin_h)*min(height[end_h],height[begin_h]) if (end_h-begin_h)*min(height[end_h],height[begin_h])>container else container
            if height[begin_h]>height[end_h]:
                end_h-=1
            else:
                begin_h+=1

                    
                
        return container
#C++版本
class Solution {
public:
    int maxArea(vector<int>& height) {
        int max_container=0;
        int begin=0,end=height.size()-1;
        if(height.size()<2){
            return max_container;
        }
        while(begin < end){
            max_container = (end-begin)*get_Min(height[begin],height[end])>max_container ?  (end-begin)*get_Min(height[begin],height[end]) : max_container;
            height[begin]>height[end]? end--:begin++;
        }
        return max_container;
    }
    int get_Min(int x,int y){
        return x<y? x:y;
    }
};