leetcode 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

12 + 92 = 82

82 + 22 = 68

62 + 82 = 100

12 + 02 + 02 = 1

思路一：使用set

#include<iostream>
#include<set>
using namespace std;

int fun(int m){
int sum = 0;
while (m){
sum += (m % 10)*(m % 10);
m /= 10;
}
cout << sum << " ";
return sum;
}

void main(){
int  n = 0;
cout << "input : "<<endl;
cin >> n;
if(n<0)
cout<<"false";
set<int> s;
set<int>::iterator ite;
int summ = fun(n);
while (true){
if (summ == 1){
cout << "true ";
break;
}
else if (s.find(summ) != s.end()){
cout << "false";
break;
}
s.insert(summ);
summ = fun(summ);
}
cin.get();
cin.get();
}

思路二：使用 Floyd Cycle detection algorithm（弗洛伊德判圈算法，也叫龟兔赛跑算法），主要思想就是如果存在圈，就想象成操场上一快一慢的两个人一起出发绕操场跑步，最终会相遇。空间复杂度O(1), 线性时间复杂度。明显优于算法一。

class Solution {
public:

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;
}
};