[LeetCode]53. Happy Number快乐数

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

• 1² + 9² = 82
• 8² + 2² = 68
• 6² + 8² = 100
• 1² + 0² + 0² = 1

Credits:
Special thanks to @mithmatt and @ts for adding this problem and creating all test cases.

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这道题定义了一种快乐数，就是说对于某一个正整数，如果对其各个位上的数字分别平方，然后再加起来得到一个新的数字，再进行同样的操作，如果最终结果变成了1，则说明是快乐数，如果一直循环但不是1的话，就不是快乐数。

 

解法1：可以发现，任意一个大于1的数执行上述循环过程若干次后，都会得到一个小于等于10的数。因此我们先算出[0,10]之间哪些是快乐数并保存下来，然后对输入数字不断进行上述过程循环，直至小于等于10，再判断即可。

class Solution {
public:
    bool isHappy(int n) {
        vector<bool> yes = {false, true, false, false, false, false, false, true, false, false, true};
        while(n > 10)
        {
            int num = 0;
            string s = to_string(n);
            for(int i = 0; i < s.size(); ++i)
                num += (int)pow(s[i] - '0', 2);
            n = num;
        }
        return yes[n];
    }
};

 

解法2：根据定义，一个数经过上述循环过程，要么得到1退出，要么无限循环下去。可以发现，无限循环下去的情况会是在若干次循环过程中，某两次得到了不为1的相同数字。因此可以将循环过程中得到的数字存下来，再下一次循环后得到的数字与前面的进行比较，若出现过则终止循环并与1进行比较。

class Solution {
public:
    bool isHappy(int n) {
        unordered_map<int, int> m;
        while(n != 1)
        {
            int num = 0;
            string s = to_string(n);
            for(int i = 0; i < s.size(); ++i)
                num += (int)pow(s[i] - '0', 2);
            n = num;
            if(m.find(n) != m.end()) break;
            else ++m[n];
        }
        return n == 1;
    }
};