题目:最大的盛水容器
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.
原题链接地址:https://leetcode.com/problems/container-with-most-water/
分析:该题难度不大,题意是给定一个int型数字num,判断其是否为回文数字。主要有两种情况:
1. num >= 0, 当num = 00000, 含多个零时,要单独讨论;
2. num < 0, 先转为正数, 再进行判断。
Java 代码:(accepted)
public class ContainerWithMostWater {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] a = { 1, 4, 5, 6, 3, 8, 9, 12, 45 };
System.out.println("Max Area is : " + maxArea(a));
int[] a1 = { 12,45,32,78,11,23,14 };
System.out.println("Max Area is : " + maxArea(a1));
int[] a2 = { 90,34,45,21,43,32};
System.out.println("Max Area is : " + maxArea(a2));
}
public static int maxArea(int[] height) {
int maxArea = 0; // The max area
int leftHight = 0;
int rigtHight = height.length - 1;
while (leftHight < rigtHight) {
//Calculate the max area
maxArea = Math.max(maxArea, (rigtHight - leftHight)
* Math.min(height[leftHight], height[rigtHight]));
//Because of cast effect, hence choose shortest height
if (height[leftHight] > height[rigtHight])
rigtHight--;
else
leftHight++;
//System.out.println("Left: " + height[leftHight] + " " + leftHight + " Right: " + height[rigtHight] + " " + rigtHight);
//System.out.println("Max Area: " + maxArea);
}
return maxArea;
}
}
测试结果:
Max Area is : 30
Max Area is : 92
Max Area is : 172
相关代码放在个人github:https://github.com/gannyee/LeetCode/tree/master/src
版权声明:本文为博主原创文章,未经博主允许不得转载。
LeetCode解题报告--Container With Most Water
原文地址:http://blog.csdn.net/github_27609763/article/details/47616077