【LeetCode】202. Happy Number

题目：

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.

Example: 19 is a happy number

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

比较直接的想法就是用一个map建立一个hash_table，对应的值为1说明出现过，那就是false。排除掉n==1的情况。不过耗时9ms，效率太低。

class Solution {
public:
    bool isHappy(int n) {
        bool happy = false;
        unordered_map<int, int> mps;
        int tmpNum = n;

        if (n == 1) return true;
        while (tmpNum != 1) {
            int buff = 0;
            mps[tmpNum] = 1;
            while (tmpNum % 10 != 0 || tmpNum / 10 != 0 ) {
                buff += (tmpNum % 10)*(tmpNum % 10);
                tmpNum /= 10;
            }
            tmpNum = buff;
            if (buff == 1) {
                happy = true;
                break;
            }
            else if (mps[buff] == 1)        break;
        }
        return happy;
    }
};

参考discussion的别人的代码，只用了4ms

class Solution {
public:
    bool isHappy(int n) {
        int sum = 0;

        if (n == 1) return true;
        if (n < 10 && n != 7) return false;
        
        while (n>=10) {
            sum += (n % 10)*(n % 10);
            n /= 10;
        }
        sum += n*n;
        if (sum == 1) {
            return  true;
        }
        else return isHappy(sum);
    }
};

利用递归来判断。对于小于10的来说，除了7以外都不是happy number