11. Container With Most Water

题目：

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.

链接：http://leetcode.com/problems/container-with-most-water/

题解：Two Pointers, 每次移动至较小的一个pointer。Time Complexity - O(n)， Space Complexity - O(1)

public class Solution {
public int maxArea(List<Integer> height) {
if(height == null || height.size() < 2)
return 0;
int max = 0, left = 0, right = height.size() - 1;

while(left < right){
int curArea = Math.min(height.get(left), height.get(right)) * (right - left);
max = Math.max(max, curArea);
if(height.get(left) < height.get(right))
left++;
else
right--;
}

return max;
}
}

二刷：

经典的Two Pointers方法，前后夹逼一下就可以得到结果了。二刷时发现signature变了唉。

Java:

public class Solution {
public int maxArea(int[] height) {
if (height == null || height.length == 0) {
return 0;
}
int maxVolumn = 0;
int lo = 0, hi = height.length - 1;
while (lo < hi) {
int curVolumn = Math.min(height[lo], height[hi]) * (hi - lo);
maxVolumn = Math.max(maxVolumn, curVolumn);
if (height[lo] < height[hi]) {
lo++;
} else {
hi--;
}
}
return maxVolumn;
}
}

Python:

class Solution(object):
def maxArea(self, height):
"""
:type height: List[int]
:rtype: int
"""
lo = 0
hi = len(height) - 1
maxVolumn = 0
while lo < hi:
curVolumn = min(height[lo], height[hi]) * (hi - lo)
maxVolumn = max(maxVolumn, curVolumn)
if height[lo] < height[hi]:
lo += 1
else:
hi -= 1
return maxVolumn

题外话：

Python真的短，希望深入学习以后能更短。

三刷:

这种小的code snippet一定要熟练，说不定可以用到哪一个大段的程序里。 像这道题的方法，就可以在"Trapping Rain Water"里面使用。很多算法，思想都是想通的，要多学多练，融会贯通。

这里我们的面积等于左边和右边较小的bar * (hi - lo)，每次都要尝试更新一下max。 之后，移动两根bar中较短的一根。

Java:

Time Complexity - O(n)， Space Complexity - O(1)

public class Solution {
public int maxArea(int[] height) {
if (height == null || height.length < 2) return 0;
int lo = 0, hi = height.length - 1;
int max = 0;
while (lo < hi) {
max = Math.max(Math.min(height[lo], height[hi]) * (hi - lo), max);
if (height[lo] < height[hi]) lo++;
else hi--;
}
return max;
}
}

Reference:
https://leetcode.com/discuss/1074/anyone-who-has-a-o-n-algorithm