LeetCode Happy Number高效解法

// Date : 2016.08.04
// Author : yqtao
// https://github.com/yqtaowhu

/************************************************************************
*
* 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.
*
************************************************************************/
//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 aboveclass 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[n]=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;    }};