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# 11.Container-With-Most-Water

## 题目描述

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 and n is at least 2.
Example:
Input: [1,8,6,2,5,4,8,3,7]
Output: 49

## 代码

### Approach 1: Brute Force

class Solution {
public int maxArea(int[] height) {
int maxarea = 0;
for (int i = 0; i < height.length; i++) {
for (int j = i + 1; j < height.length; j++) {
maxarea = Math.max(maxarea, Math.min(height[i], height[j]) * (j - i));
}
}
}
}

### Approach #2 Two Pointer Approach

Complexity Analysis
• Time complexity : O_(_n). Single pass.
• Space complexity : O(1). Constant space is used.
class Solution {
public int maxArea(int[] height) {
int maxarea = 0, l = 0, r = height.length - 1;
while (l < r) {
maxarea = Math.max(maxarea, Math.min(height[l], height[r]) * (r - l));
if (height[l] < height[r]) {
l++;
} else {
r--;
}
}
return maxarea;
}
}