LeetCode Happy Number高效解法

// Date : 2016.08.04

// Author : yqtao

// https://github.com/yqtaowhu

/[b]************************************************************************[/b]

*

* 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

*

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

*

[b]************************************************************************[/b]/

//very quikly ,just using 0 ms

//reference :https://en.wikipedia.org/wiki/Happy_number

//just know 1 is happy number,and 4 is not

class Solution {
public:
bool isHappy(int n) {
int num=0;
while(n!=1&&n!=4) {
while(n) {
num += (n%10) * (n%10);
n/=10;
}
n=num;num=0;
}
return 1==n;
}
};

另一种递归解法，与上述思想是一致的

//recursion , the idea is simple to above
class Solution {
public:
bool isHappy(int n) {
if (n==1) return true;
if (n==4) return false;
int num=0;
while (n) {
int t=n%10;
num+=t*t;
n/=10;
}
return isHappy(num);
}
};

最后一种解法，普遍的解法

//using map
//if you dont know 4 is unhappy,just using it
class Solution {
public:
bool isHappy(int n) {
int num=0;
unordered_map<int,bool> table;
table
=1;
while(n!=1)
{
while(n)
{
num += (n%10) * (n%10);
n/=10;
}
if(table[num]) break; //is equal to pre number ,break ,if not do ,it will always iterator.
else table[num]=1;    //this is why using a map to control.
n=num;num=0;
}
return 1==n;
}
};