LeetCode 11. Container With Most Water

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本文介绍了一个经典的两指针算法问题：给定一系列垂直线，每条线的长度由数组给出，求能够组成的容器中能盛放的最大水量。通过双指针技巧，逐步移动较短的边界，有效地解决了这个问题。

Given n non-negative integers a₁, a₂, ..., a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) 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.

Classical two pointers problem

#include <vector>
#include <climits>
#include <iostream>
using namespace std;

int containMostWater(vector<int>& height) {
    if(height.size() <= 1) return 0;
    int i = 0;
    int j = height.size() - 1;
    int maxWater = 0;
    while(i < j) {
        int minHeight = min(height[i], height[j]);
        maxWater = max(maxWater, minHeight * (j - i));
        if(height[i] > height[j]) j--;
        else i++;
    }
    return maxWater;
}

int main(void) {
    vector<int> height{2, 3, 4, 3, 2};
    int maxWater = containMostWater(height);
    cout << maxWater << endl;
}