• 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: 

    Input: 19
    Output: true
    Explanation: 
    12 + 92 = 82
    82 + 22 = 68
    62 + 82 = 100
    12 + 02 + 02 = 1

    Approach #1: C++.

    class Solution {
    public:
        bool isHappy(int n) {
            unordered_set<int> s;
            s.insert(n);
            while (1) {
                int summ = 0;
                while (n) {
                    int t = n % 10;
                    summ += t*t;
                    n /= 10;
                }
                if (summ == 1) return true;
                if (s.count(summ)) return false;
                s.insert(summ);
                n = summ;
            }
        }
    };
    

      

    Approach #2: Java.

    class Solution {
        public boolean isHappy(int n) {
            HashSet<Integer> set = new HashSet<Integer>();
            set.add(n);
            while (n != 1) {
                int result = 0;
                while (n != 0) {
                    result += Math.pow(n % 10, 2);
                    n /= 10;
                }
                if (set.contains(result)) {
                    return false;
                }
                set.add(result);
                n = result;
            }
            return true;
        }
    }
    

      

    Approach #3: Python.

    class Solution(object):
        def isHappy(self, n):
            """
            :type n: int
            :rtype: bool
            """
            mem = set()
            while n != 1:
                n = sum([int(i)**2 for i in str(n)])
                if n in mem:
                    return False
                else:
                    mem.add(n)
            else:
                return True
    

      

    Time SubmittedStatusRuntimeLanguage
    a few seconds ago Accepted 28 ms python
    2 minutes ago Accepted 5 ms java
    8 minutes ago Accepted 0 ms cpp
    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/9959997.html
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