202. Happy Number

问题描述

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.

题目链接：

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思路分析

将一个数的每一位的平方加和起来，不断重复，如果最后的到1，那这个数就是happy number。判断一个数是不是happy number。

设置循环10次，每次迭代计算置换后的值，如果位1就true，否则10次之后返回false。

代码

class Solution {
public:
    bool isHappy(int n) {
        int count = 10;
        while(count){
            n = replace(n);
            if (n == 1)
                return true;
            count--;
        }
        return false;
    }

    int replace(int n){
        int result = 0;
        while(n != 0){
            result += pow(n%10, 2);
            n = n/10;
        }
        return result;
    }
};

时间复杂度：$O(logn)$
空间复杂度：$O(1)$

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反思

这个方法竟然可以，也是瞎猫撞上死耗子了。主要需要优化的地方在于如何判断这数肯定不是happy的了。

class Solution {
public:
    bool isHappy(int n) {
        int fast, slow;
        fast = slow = n;
        do{
            slow = replace(slow);
            fast = replace(fast);
            fast = replace(fast);
            if (fast == 1)
                return true;
        }while (slow != fast);
            return false;
    }

    int replace(int n){
        int result = 0;
        while(n != 0){
            result += pow(n%10, 2);
            n = n/10;
        }
        return result;
    }
};

时间复杂度：$O(n)$
空间复杂度：$O(1)$

在Discussion区发现的惊为天人的做法，使用两个指针，fast指针每个循环多计算一遍，比slow的计算速度要快。如果slow与fast相同了，那表明要么n稳定在1了；要么存在一个环，就像判断两条链表在哪了相交一样，fast会追上slow。