LeetCode day6-优快云博客

本文链接：https://blog.youkuaiyun.com/Shui_De_Tian_Ya/article/details/108214646

本文介绍了一种判断正整数是否为快乐数的算法。快乐数是指通过将该数替换为其各位数字的平方和，重复此过程直到数字变为1或进入不包含1的循环。文章提供了一个使用快慢指针技巧的C++实现，用于检测循环并确定最终结果是否为1。

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202.Happy Number

Write an algorithm to determine if a number n is “happy”.

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.

Return True if n is a happy number, and False if not.

Example:

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分析

构造一个求一个正整数的各位平方和的函数，已知是一个循环过程，所以可以运用快慢标识来计算，当快慢标识的数相等时，判断值是否为1，是的话输出true，否则输出false。

代码

#include<iostream>
using namespace std;

class Solution {
public:

	int getNum(int n) {
		int sum = 0;
		while(n) {
			sum += (n % 10)*(n % 10);
			n = n / 10;
		}
		return sum;
	}

	bool isHappy(int n) {
		int slow = n;
		int fast = n;
		do {
			slow = getNum(slow);
			fast = getNum(fast);
			fast = getNum(fast);
		} while (slow != fast);

		return slow == 1 ? true : false;
	}
	
};

int main() {
	int i;
	Solution s;
	for (i = 1; i < 100; i++) {
		if (s.isHappy(i))
			printf("%d ", i);
	}
	return 0;
}