Container With Most Water 383

Question

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.

Notice

You may not slant the container.

Example

Given [1,3,2], the max area of the container is 2.

Solution

这题和Trapping rain water不一样，取任意两个木板后，中间的木板可以忽略。

首先想到的是两次循环找最大值，时间复杂度O(n^2)。

如果用前后指针则可以将时间复杂度降到O(n)。计算取前后指针为木板时的面积，然后移动较短的那一个。保留最大面积。

代码如下：

O(n^2)

public class Solution {
    /**
     * @param heights: an array of integers
     * @return: an integer
     */
    public int maxArea(int[] heights) {
        // write your code here
        if(heights == null || heights.length <= 1){
            return 0;
        }
         int area = 0;
         for(int i = 0; i < heights.length - 1; i++){
             for(int j = i + 1; j < heights.length; j++){
                 int width = j - i;
                 int height = Math.min(heights[i], heights[j]);
                 area = Math.max(area, width * height);
             }
         }
         return area;
    }
}

O(n)

public class Solution {
    /**
     * @param heights: an array of integers
     * @return: an integer
     */
    public int maxArea(int[] heights) {
        // write your code here
        if(heights == null || heights.length <= 1){
            return 0;
        }

        int start = 0;
        int end = heights.length - 1;
        int sum = 0;
        while(start < end){
            sum = Math.max(sum, (end - start) * Math.min(heights[start], heights[end]));
            if(heights[start] < heights[end]){
                start++;
            }else{
                end--;
            }
        }
        return sum;
    }
}