Leet Code 11 Container With Most Water

最新推荐文章于 2023-10-13 09:04:47 发布

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最新推荐文章于 2023-10-13 09:04:47 发布

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分类专栏： Leet Code

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本文探讨了如何通过双指针技巧求解容器最大容量问题，详细解释了算法原理及其实现过程，提供了完整的Java代码示例。

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.

思路：利用一个性质，ai右边更小的边ak (k > i) 与 aj (j > k) 之间的面积一定比ai,aj围成的面积更小。

 public int maxArea(int[] height) {
        if(height.length <= 1)
        {
            return 0;
        }
        
        int left = 0;
        int right = height.length - 1;
        int maxArea = right * ( height[left] < height[right] ? height[left] : height[right]);
        int tmpArea = 0;
        
        while(left < right)
        {
            tmpArea = (right - left) * ( height[left] < height[right] ? height[left] : height[right]);
            maxArea = tmpArea > maxArea ? tmpArea : maxArea;
            
            if(height[left] < height[right])
            {
                int k = left;
                while((k < right) && (height[k] <= height[left]))
                {
                    k ++;
                }
                left = k;
            }
            
            else
            {
                int k = right;
                while((k > left) && (height[k] <= height[right]))
                {
                    k --;
                }
                right = k;
            }
        }
        
        return maxArea;
    }