leetcode---C++实现---11. Container With Most Water（盛最多水的容器）

题目

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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/container-with-most-water

解题思路

本题想要计算区间上能组成的最大面积，两条边的组成为：两条vertical line中较矮的高度，两条vertical line的水平距离。
当两条vertical line中有一条较高时：
若将较高的那条line往里移，在较高line的高度不变时，移动后的两条line的水平距离缩小了，只能比原先的距离更小，因此该情况不进行计算考虑；
若固定较高line，将较短的line往里移，有可能遇到更高的line，有可能出现更大的面积，此情况需要计算，在所有计算的面积中选出最大面积即可。

算法实现

#include <algorithm>
class Solution {
public:
    int maxArea(vector<int>& height) {
        int left = 0;
        int right = height.size() - 1;
        int area = 0;
        while (left < right)
        {
            int curArea = min(height[left], height[right]) * (right - left);
            area = max(curArea, area);
            if (height[left] < height[right])    
                ++left;
            else    
                --right;
        }

        return area;
    }
};