[leetcode]11. Container With Most Water存水最多的容器

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 and n is at least 2.

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

题意：

给定一堆高矮不一的平行的墙，选择其中两堵作为两壁，问最多能存多少水。

Solution1：

Two Pointers:  set pointer left at index 0; set pointer right at index len - 1

We calculate area base on width (gap of left and right index ) and height(smaller height matters because of "Short-board effect" )

In order to find max area, we move the pointer whichever representing the smaller height

code:

 1 /*
 2     Time complexity : O(n). We traverse the whole give array
 3     Space complexity : O(1). We only used constant extra space.
 4 */
 5
 6 class Solution {
 7     public int maxArea(int[] height) {
 8         int left = 0;
 9         int right = height.length - 1;
10         int area = Integer.MIN_VALUE;
11         while(left < right){
12             area = Math.max(area, Math.min(height[left], height[right]) * (right - left));
13             if(height[left] < height[right]){
14                 left ++;
15             }else{
16                 right --;
17             }
18         }
19       return area;
20     }
21 }

原文地址：https://www.cnblogs.com/liuliu5151/p/10657814.html