Container With Most Water

最新推荐文章于 2024-08-28 08:50:09 发布

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本文介绍了一种寻找能盛最多水的两个垂直线的高效算法。给定一系列非负整数，代表坐标上的高度，通过计算不同组合下由两线与x轴构成的容器盛水量，找出最大的盛水量。采用双指针法，复杂度为O(N)，并详细解释了算法思路。

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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 linei 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：(j-i)*Math.min(height[i],height[j]); Since i is lower than j, so there will be no jj < j that make the area from i,jj is greater than area from i,j. so the maximum area that can benefit from i is already recorded.thus, we move i forward. 这个思路很巧妙。O(N)

class Solution {
    public int maxArea(int[] height) {
        if(height == null || height.length == 0) {
            return 0;
        }
        int i = 0; int j = height.length - 1;
        int maxres = 0;
        while(i < j) {
            maxres = Math.max(maxres, (j - i) * Math.min(height[i], height[j]));
            if(height[i] < height[j]) {
                i++;
            } else {
                // height[i] > height[j];
                j--;
            }
        }
        return maxres;
    }
}