Leetcode 11. Container With Most Water , Java解法

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.

 

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

Approach #1 Brute Force 我的解法

public class Solution {
    public int maxArea(int[] height) {
        int max = 0;
        int len = height.length;
        // try every possible pair
        for(int i=0;i<len;i++){
            for(int j=i+1;j<len;j++){
                max = Math.max(max,Math.min(height[i],height[j])*(j-i));
            }
        }
        return max;
    }
}

Complexity Analysis

• Time complexity : O(n^2). Calculating area for all \dfrac{n(n-1)}{2}​ height pairs.
• Space complexity : O(1). Constant extra space is used. 

Approach #2: Two Pointer 

public 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;
    }
}

从Array的两边分别开始计算乘积（距离，和两端更矮的一端的高度），然后更新最大值，移动更矮一端的坐标。

Complexity Analysis

• Time complexity : O(n) Single pass.

• Space complexity : O(1). Constant space is used.