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。

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。

注意这里的高度是两根bar中较小的一根，长度是 hi - lo。可以想象成两根细线中间夹的区域，并不是hi - lo + 1。

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