202. Happy Number

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

首先想到的比较简单 就按照题目叙述来 用set防重 

private HashSet<Integer> set = new HashSet<Integer>();

public boolean isHappy(int n) {
    if(n == 1) return true;
    int temp, result = 0;
    while(n > 0){
        temp = n%10;
        result += temp * temp;
        n = n/10;
    }
    if(set.contains(result)) return false;
    set.add(result);
    return isHappy(result);
}

空间复杂度O(n)

但是discuss里惊现神操作

I see the majority of those posts use hashset to record values. Actually, we can simply adapt the Floyd Cycle detection algorithm. I believe that many people have seen this in the Linked List Cycle detection problem. The following is my code:

int digitSquareSum(int n) {
    int sum = 0, tmp;
    while (n) {
        tmp = n % 10;
        sum += tmp * tmp;
        n /= 10;
    }
    return sum;
}

bool isHappy(int n) {
    int slow, fast;
    slow = fast = n;
    do {
        slow = digitSquareSum(slow);
        fast = digitSquareSum(fast);
        fast = digitSquareSum(fast);
    } while(slow != fast);
    if (slow == 1) return 1;
    else return 0;
}

想法非常像判断链表中是否有环

就想题目中叙述的 这种计算一定会陷入循环 如果循环时=1 那就是happy num 否则不是 

空间复杂度O(1)