Container with Most Water

最新推荐文章于 2024-08-28 08:50:09 发布

转载最新推荐文章于 2024-08-28 08:50:09 发布 · 57 阅读

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原文链接：https://segmentfault.com/a/1190000008124293

本文介绍了一种解决“容器盛水”问题的有效算法。通过使用双指针技术，该算法能够在O(n)的时间复杂度内找到能容纳最多水的两个线段。文中详细解释了算法的思路，并给出了具体的Java代码实现。

http://www.lintcode.com/en/pr...

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.

**思路**
对撞指针问题， Volume = Width * Min(left_Height, right_Height). 求最大体积，当缩小宽度时， 则高度必须比原来大。---->两边指针选较小的一个靠近直到比原来的Height大。 # 此程序实现中省略了内层while。 



public class Solution {
    public int maxArea(int[] height) {
        if (height == null || height.length < 2){
            return 0;
        }
        int res = 0;
        int left = 0, right = height.length -1;
        while (left < right){
            int minHeight = Math.min(height[left], height[right]);
            res = Math.max(res, minHeight * (right - left));
            if (height[left] < height[right]){
                left ++;
            }
            else{
                right --;
            }
        }
        return res;
    }
}