LeetCode 11. Container With Most Water（容器装水）

原题网址：https://leetcode.com/problems/container-with-most-water/

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.
方法一：构造一个递增容器序列，当遇到高度降低时，检查左边所有可能的容器。

public class Solution {
public int maxArea(int[] height) {
List<Integer> list = new ArrayList<>();
int max = 0;
int area = 0;
for(int i=0; i<height.length; i++) {
if (list.isEmpty() || height[list.get(list.size()-1)] < height[i]) {
list.add(i);
} else {
for(int j=0; j<list.size(); j++) {
area = (i-list.get(j)) * Math.min(height[list.get(j)], height[i]);
if (area > max) max = area;
if (height[list.get(j)] > height[i]) break;
}
}
}
for(int j=0; j<list.size()-1; j++) {
area = (list.get(list.size()-1) - list.get(j)) * height[list.get(j)];
if (area > max) max = area;
}
return max;
}
}

方法二：从左右两边往中间扫描，当其中一边较矮的时候，及时计算该侧所能装的水，然后该侧内移。

public class Solution {
public int maxArea(int[] height) {
int max = 0;
int left = 0, right = height.length-1;
while (left < right) {
if (height[left] < height[right]) {
int area = height[left] * (right-left);
if (area > max) max = area;
left ++;
} else if (height[left] > height[right]) {
int area = height[right] * (right-left);
if (area > max) max = area;
right --;
} else {
int area = height[left] * (right-left);
if (area > max) max = area;
left ++;
right --;
}
}
return max;
}
}

方法一和二的思路：