【LeetCode】解题11：Container With Most Water

Problem 11: Container With Most Water [Medium]

Given n non-negative integers a₁, a₂, …, a_n , where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) 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

来源：LeetCode

解题思路

1. 暴力求解

使用二重循环，遍历$\frac{n(n-1)}{2}$对组合的容量，返回最大值。
时间复杂度O($n^2$)。

2. 双指针

使用双指针指向数组的头和尾，并在循环中不断将指向较短线段的指针向内移动，同时不断更新最大容量的数值，直到两个指针指向同一位置。
移动原理： 初始指针的位置在x轴上的长度是最长的，但将指向较短线段的指针向内移动时，虽然x轴上的长度变短，却有可能因为指向的线段变长而使总容量变长。
整个过程只对数组进行了一次遍历，因此时间复杂度为O($n$)。

Solution (Java)

双重循环：

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

双指针：

class Solution {
    public int maxArea(int[] height) {
        int N = height.length;
        int max = 0;
        int left = 0, right = N-1;
        while(left != right){
            max = Math.max(max, Math.min(height[left], height[right]) * (right - left));
            if(height[left] < height[right]) left += 1;
            else right -= 1;
        }
        return max;
    }
}