C#LeetCode刷题之#202-快乐数(Happy Number)

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!

C#LeetCode刷题之#202-快乐数(Happy Number)

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