LeetCode. Container With Most Water

试题概述

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.

题目大概是说在笛卡尔坐标系中给定n个点, 这n个点的坐标都在x轴上方, 向x轴作垂线段可得到x条线段. 从中选择两条线段, 使得这两条线段和x轴形成的水缸得面积最大.

解题思路

最简单粗暴的方法是穷举这n22种组合方式, 不用想也知道妥妥的TLE.

假设有2个指针i, j. i从最左侧出发, j从最右侧出发. 因为水缸的高度是由二者中的较小值决定的. 当height[i] < height[j]时, ++ i. 假设i的值不变, – j, 不可能得到更优解(因为水缸的长度在减小, 高度不可能超过height[i]). 反之, – j. 并记录每次移动i, j后的水缸面积, 并更新水缸面积的最大值.

源代码

public class Solution {    public int maxArea(int[] height) {        int i = 0, j = height.length - 1;        int maxArea = getArea(height, i, j);        while ( i < j ) {            if ( height[i] < height[j] ) {                ++ i;            } else {                -- j;            }            int currentArea = getArea(height, i, j);            if ( currentArea > maxArea ) {                maxArea = currentArea;            }        }        return maxArea;    }    private int getArea(int[] height, int i, int j) {        int a = Math.min(height[i], height[j]);        int b = j - i;        return a * b;    }    public static void main(String[] args) {        Solution s = new Solution();        int[] height = {2, 3, 10, 5, 7, 8, 9};        System.out.println(s.maxArea(height));    }}