LeetCode-11. Container With Most Water

最新推荐文章于 2018-12-20 15:38:20 发布

Alatebloomer

最新推荐文章于 2018-12-20 15:38:20 发布

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分类专栏： leetcode/Array

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本文解决LeetCode-11盛水容器问题，介绍如何通过双指针法找到能容纳最多水的两个垂直线。初始时，两指针分别位于线段两端，逐步向中间靠拢并更新最大容量。

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LeetCode-11. Container With Most Water

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.

解题思路：此题要求求解两条垂直线与X轴围成的容器乘放的水量，将两端的点看成是宽度最大的容器，之后的容器宽度不会比这个大，然后依次减小宽度，遇到高度不大于原来的高度的线则跳过，比较新容器与之前的最大乘水量。

class Solution {
public:
    int maxArea(vector<int>& height) {
        int i = 0, j = height.size() - 1;
		int water = 0;
		while (i < j)
		{
			int h = min(height[i], height[j]);
			water = max(water, (j - i)*h);
			while (height[i] <= h && i < j)++i;
			while (height[j] <= h && i < j)--j;
		}
		return water;
        
    }
};