一、题目叙述:
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.
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二、解题思路:
这个我真的没想出来,头脑简单的用暴力写,显然复杂度到n2了,就超时了,最后也没想出来怎么弄。所以看了别人的做法,结果非常简单,复杂度也就O(N)。基本思路是这样的:在数组前后各放一个指针,计算此时容器的能盛水的量,因为显然此时的矩形长达到了最长,接着将短的一边的指针向前移,向前移时,矩形长已经在缩短了,只要想办法提升宽面积才能增大,由此。。。
三、源源源码:
public class Solution { public int maxArea(int[] height) { int min = 0; int max = height.length - 1; int maxarea = 0; int temp = 0; while (min < max) { temp = (max - min) * Math.min(height[min], height[max]); if (temp > maxarea) maxarea = temp; if (height[min] < height[max]) min ++; else max --; } return maxarea; } public static void main(String args[]) { int[] a = {1,1}; Solution solution = new Solution(); System.out.println(solution.maxArea(a)); } }