Happy Number -- leetcode

标签：leetcode   平方和   快慢指针   

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

算法一，

既然有死循环的可能，使用一个集合，记录已经访问过的数字。如果在集合中出现，则出现了循环。

在leetcode上用时8ms。

class Solution {
public:
    bool isHappy(int n) {
        unordered_set<int> visited;
        while (n != 1 && !visited.count(n)) {
            visited.insert(n);
            int number = 0;
            while (n) {
                number += (n % 10) * (n % 10);
                n /= 10;
            }
            n = number;
        }
        return  n == 1;
    }
};

算法二，快慢指针

使用集合到底会额外占空间。

此处使用快慢指针的概念，来探知是否存在死循环。

在leetcode上实际执行时间为4ms。

class Solution {
public:
    bool isHappy(int n) {
        int slow = n, fast = n;
        do {
            slow = next(slow);
            fast = next(next(fast));
        } while (slow != fast);
        return slow == 1;
    }
    
    int next(int n) {
        int ans = 0;
        while (n) {
            ans += (n % 10) * (n % 10);
            n /= 10;
        }
        return ans;
    }
};

Happy Number -- leetcode

标签：leetcode   平方和   快慢指针   

原文地址：http://blog.csdn.net/elton_xiao/article/details/46845563