C#LeetCode刷题之#202-快乐数（Happy Number）

问题

编写一个算法来判断一个数是不是“快乐数”。

一个“快乐数”定义为：对于一个正整数，每一次将该数替换为它每个位置上的数字的平方和，然后重复这个过程直到这个数变为 1，也可能是无限循环但始终变不到 1。如果可以变为 1，那么这个数就是快乐数。

    输入: 19

    输出: true

    解释:

    1^{2} + 9^{2} = 82
    8^{2} + 2^{2} = 68
    6^{2} + 8^{2} = 100
    1^{2} + 0^{2} + 0^{2} = 1

Write an algorithm to determine if a number 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.

    Input: 19

    Output: true

    Explanation:

    1^{2} + 9^{2} = 82
    8^{2} + 2^{2} = 68
    6^{2} + 8^{2} = 100
    1^{2} + 0^{2} + 0^{2} = 1

示例

    public class Program {
     
        public static void Main(string[] args) {
            int nums = 19;
     
            var res = IsHappy(nums);
            Console.WriteLine(res);
     
            nums = 26;
            res = IsHappy2(nums);
            Console.WriteLine(res);
     
            nums = 65;
            res = IsHappy3(nums);
            Console.WriteLine(res);
     
            Console.ReadKey();
        }
     
        private static bool IsHappy(int n) {
            //递归法
            var set = new HashSet<int>();
            return ComputerHappy(n, set);
        }
     
        private static bool ComputerHappy(int n, HashSet<int> set) {
            if(n == 1) return true;
            if(set.Contains(n)) return false;
            set.Add(n);
            //计算各位数平方和
            //int sum = 0;
            //for(int i = 0; i < n.ToString().Length; i++) {
            //    sum += (int)Math.Pow(n / (int)Math.Pow(10, i) % 10, 2);
            //}
            return ComputerHappy(SquareSum(n), set);
        }
     
        private static int SquareSum(int n) {
            //计算各位数平方和
            var sum = 0;
            while(n > 0) {
                sum += (n % 10) * (n % 10);
                n = n / 10;
            }
            return sum;
        }
     
        private static bool IsHappy2(int n) {
            //若一个数不是快乐数，其必定死在4手上，最终进入以下死循环
            //4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4
            //4的各数位平方是16，将y进行2次各位数平方和计算
            //若相等，则到达了4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4
            //这样的死循环或到达了1这样的快乐数结束值
            //返回x == 1即可判定是不是快乐数
            var x = n;
            var y = n;
            do {
                x = SquareSum(x);
                y = SquareSum(SquareSum(y));
            } while(x != y);
            return x == 1;
        }
     
        private static bool IsHappy3(int n) {
            //参考IsHappy2
            while(n != 1 && n != 4) {
                var sum = 0;
                while(n > 0) {
                    sum += (n % 10) * (n % 10);
                    n /= 10;
                }
                n = sum;
            }
            return n == 1;
        }
     
    }

以上给出3种算法实现，以下是这个案例的输出结果：

    True
    False
    False

分析：

该题的时间复杂度分析依赖于一些数学知识，若有看官知道如何分析请给我留言，不胜感激！
---------------------
作者：无痕的过往
来源：CSDN
原文：https://blog.csdn.net/qq_31116753/article/details/82598172
版权声明：本文为博主原创文章，转载请附上博文链接！