Skip to content

Create Felix.md #3

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
merged 1 commit into from
Nov 26, 2018
Merged

Create Felix.md #3

Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
30 changes: 30 additions & 0 deletions 2019.11.25-leetcode80/Felix.md
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,30 @@
```javascript
package leetcode.medium;
/**
* @author Felix
* @date 2018年10月31日下午9:59:27
@version 给定 n 个非负整数 a1，a2，...，an，每个数代表坐标中的一个点 (i, ai) 。在坐标内画 n 条垂直线，垂直线 i 的两个端点分别为
(i, ai) 和 (i, 0)。找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。说明：你不能倾斜容器，且 n 的值至少为 2。
*/
public class MaxArea {
public int maxArea(int[] height){
int len = height.length;
int max = 0,low = 0,high = len - 1;
while(low < high){
int lowMax = height[low],highMax = height[high];
int water = (high - low) * (lowMax < highMax ? lowMax : highMax);
max = water > max ? water : max;

if(height[low] <= height[high]){
while(low < high && height[low] <= lowMax)
low++;
}else{
while(high > low && height[high] <= highMax)
high--;
}
}

return max;
}
}
```