11 Container With Most Water

Medium

Given n non-negative integers a1, a2, …, an , where each represents a point at coordinate (i, ai). nvertical 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:

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

给定 n 个非负整数 a1，a2，…，an，每个数代表坐标中的一个点 (i, ai) 。在坐标内画 n 条垂直线，垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0)。找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。

说明：你不能倾斜容器，且 n 的值至少为 2。

图中垂直线代表输入数组 [1,8,6,2,5,4,8,3,7]。在此情况下，容器能够容纳水（表示为蓝色部分）的最大值为 49。

示例:

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输入: [1,8,6,2,5,4,8,3,7]
输出: 49

想法

参考Solution做的……具体就是用双指针来扫描，从两端开始，如果$a_{left} < a_{right}$则使左指针+1，若$a_{left} > a_{right}​$则使右指针-1。证明在Solution里有……

解

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#include <iostream>
#include <vector>

using namespace std;

class Solution {
  public:
    int maxArea(vector<int> &height)
    {
        int l = 0, r = height.size() - 1;
        int maxValue = 0;
        while (l < r) {
            maxValue = max(maxValue, (r - l) * min(height[l], height[r]));
            if (height[l] < height[r]) {
                l++;
            }
            else {
                r--;
            }
        }
        return maxValue;
    }
};

int main(void)
{
    vector<int> v({1, 8, 6, 2, 5, 4, 8, 3, 7});
    Solution s;
    cout << s.maxArea(v) << endl;
    return 0;
}