Algorithm——Container With Most Water

Q:

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

A:

JAVASCRIPT（這是log(n^2)的方法，還有很大的優化區間）

/**
 * @param {number[]} height
 * @return {number}
 */
var maxArea = function(numArr) {
    var h = '', w = '', tmp = 0, max = 0;
    for (var m = 0; m < numArr.length - 1; m++) {
        for (var n = m + 1; n < numArr.length; n++) {
            h = Math.min(numArr[m], numArr[n]);
            w = n - m;
            tmp = h * w;

            if (max < tmp) {
                max = tmp;
            }          
        }
    }

    return max;         
};

 JAVA（這是log(n)的方法）

/**
 * Container With Most Water
 * 盛最多水的容器
*/

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;
    }
}