【LeetCode】202 Happy Number

最新推荐文章于 2020-12-10 10:26:49 发布

原创最新推荐文章于 2020-12-10 10:26:49 发布 · 291 阅读

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本文介绍了一种算法用于判断一个正整数是否为快乐数，快乐数定义为通过将数字替换为其各位数字的平方和并重复此过程直到得到1或进入循环的数。文中提供了两种实现方法，并解释了使用HashSet解决死循环问题的原理。

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<span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; background-color: rgb(255, 255, 255);">Write an algorithm to determine if a number is "happy".</span>

A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.

Example: 19 is a happy number

12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

法1：（Runtime: 5 ms）

public class Solution {
    public boolean isHappy(int num) {
        if (num == 0 || num == 1) {
			return true;
		}
		Set<Integer> set = new HashSet<Integer>();
		set.add(num);
		while (true) {
			int sum = 0;
			while(num != 0){
				int temp = num % 10;
				sum += temp * temp;
				num = num / 10;
			}
			if(sum == 1){
				return true;
			}
			if (!set.add(sum)) {
				return false;
			}
			num = sum;
		}
    }
}

法2：（Runtime: 9 ms）

public class Solution {
    public boolean isHappy(int num) {
        if (num == 0) {
			return true;
		}
		Set<Integer> set = new HashSet<Integer>();
		set.add(num);
		String s = num + "";
		while (Integer.parseInt(s) != 1) {
			int sum = 0;
			for (int i = 0; i < s.length(); i++) {
				int temp = Integer.parseInt(s.charAt(i) + "");
				sum += temp * temp;
			}
			if(!set.add(sum)){
				return false;
			}
			s = sum + "";
		}
		return true;
    }
}

刚开始构思时，迟迟想不到用什么办法来中断计算死循环，在网上搜索了一些资料以后，发现可以用HashSet来解决死循环问题，因为Set是不重复集合，只要发现往其中添加的数是已存在的，即可判断循环计算回到死循环节点，此时即可跳出循环。