LeetCode11. Container With Most Water最容易理解的解法

最新推荐文章于 2024-01-04 13:57:00 发布

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本文探讨了一道经典的算法题目：给定一系列非负整数，代表坐标上的点的高度，如何找出两个点与x轴构成的容器能盛放最多的水。文章通过分析问题并提供了一个C++实现的解决方案，采用双指针法逐步移动边界来寻找最优解。

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.

class Solution {
public:
    
int getArea(vector<int>& height, int left, int right)
{
	int h = height[left] > height[right] ? height[right] : height[left];
	return h*(right - left);
}

int maxArea(vector<int>& height) {
	int left = 0;
	int right = height.size() - 1;
	int max = 0;
	while (left < right) {
		int m = getArea(height, left, right);
		if (m > max){
			max = m;
		}
		if (height[left] < height[right]) {
			left++;
		}
		else {
			right--;
		}
	}
	return max;
}

};