11. Container With Most Water

最新推荐文章于 2024-10-29 13:54:07 发布

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最新推荐文章于 2024-10-29 13:54:07 发布

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原文链接：http://www.cnblogs.com/joycelee/p/5398059.html

本文介绍了一种解决两直线与x轴构成的最大盛水量问题的高效算法。通过双指针技术，逐步移动较短直线对应的指针以寻找可能的最大面积。此算法避免了不必要的遍历，将时间复杂度降低到O(n)。

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.

Similar: 42. Trapping Rain Water

 1 public class Solution {
 2     public int maxArea(int[] height) {
 3         if (height.length < 2) return 0;
 4         int l = 0, r = height.length-1;
 5         int max = 0;
 6         while (l < r) {
 7             int lomax = height[l];
 8             int himax = height[r];
 9             max = Math.max(max, (r - l) * (lomax < himax ? lomax : himax));
10             
11             if (height[l] < height[r])
12                 while (l < r && height[l] <= lomax) l++; // will change l at least one step
13             else while (l < r && height[r] <= himax) r--;
14         }
15         
16         return max;
17     }
18 }