Leetcode 202. Happy Number 快乐数

202. Happy Number 快乐数

题目描述

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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.

示例:

Example:

Input: 19
Output: true
Explanation: 
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

解答1

这道题第一种解法就是一个个求，看看最后这个数各位数平方之和能不能为1，与此同时maintain一个set，用来记录已经访问过的数，如果这个数再次出现就返回false。

代码1

class Solution {
set<int > all;
public:
    bool isHappy(int n) {

        int sum = 0;
        if(all.find(n) != all.end()) { return false;}
        all.insert(n);
        while (n != 0) {
            sum += (n % 10) * (n % 10);
            n = n / 10;
        }     
        if (sum == 1) return true;
        else { 
            
            return isHappy(sum);
            
        }
    }
};

解答2

第二种方法较为独特，我们可以把Linked List Cycle detection的思想应用到这个上面。我们maintain两个数，slow和fast。slow计算一次digitSum，fast计算两次digitSum，当slow和fast相等时查看这个值是否为1，是1就返回true，否则false。

代码2

class Solution {

public:

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

    bool isHappy(int n) {
        int slow = n; 
        int fast = n;
        do {
            slow = digitSum(slow);
            fast = digitSum(fast);
            fast = digitSum(fast);

        } while (slow != fast);
        if (slow == 1) return true;
        return false;
        
    }
};