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[LeetCode]202.Happy Number

时间:2015-04-22 22:15:41      阅读:1099      评论:0      收藏:0      [点我收藏+]

标签:leetcode   经典面试题   

题目

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

代码

/*---------------------------------------------------------
*   日期:2015-04-22
*   作者:SJF0115
*   题目: 202.Happy Number
*   网址:https://leetcode.com/problems/happy-number/
*   结果:AC
*   来源:LeetCode
*   博客:
------------------------------------------------------------*/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
    bool isHappy(int n) {
        if(n == 1){
            return true;
        }//if
        if(n == 0){
            return false;
        }//if
        return Recursion(n);
    }
private:
    vector<int> visited;
    bool Recursion(int n){
        if(n == 1){
            return true;
        }//if
        int num = n;
        int tmp,value = 0;
        while(num){
            tmp = num % 10;
            value += tmp * tmp;
            num /= 10;
        }//while
        // 判断是否形成圈
        vector<int>::iterator result = find(visited.begin(),visited.end(),value);
        if(result != visited.end()){
            return false;
        }//if
        visited.push_back(value);
        return Recursion(value);
    }
};

int main(){
    Solution solution;
    int n = 2;
    if(solution.isHappy(n)){
        cout<<n<<" is Happy Number"<<endl;
    }//if
    else{
        cout<<n<<" is not Happy Number"<<endl;
    }
    return 0;
}

运行时间

技术分享

[LeetCode]202.Happy Number

标签:leetcode   经典面试题   

原文地址:http://blog.csdn.net/sunnyyoona/article/details/45201103

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