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happy-number

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Happy Number – LeetCode 202

Problem

Description

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

Example: 19 is a happy number

$1^2 + 9^2 = 82$
$8^2 + 2^2 = 68$
$6^2 + 8^2 = 100$
$1^2 + 0^2 + 0^2 = 1$

Answer

Original

Code

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class Solution {
public:
    int digitSquareSum(int n) {
        int sum = 0, tmp;
        while (n) {
            tmp = n % 10;
            sum += tmp * tmp;
            n /= 10;
        }
        return sum;
}

bool isHappy(int n) {
        int slow, fast;
        slow = fast = n;
        do {
            slow = digitSquareSum(slow);
            fast = digitSquareSum(fast);
            fast = digitSquareSum(fast);
        } while(slow != fast);
        if (slow == 1) return 1;
        else return 0;
    }
};

思路

因为非快乐数会进入一个循环，所以和昨天一样使用快慢双指针来检测。时间复杂度$O(n)$，空间复杂度$O(1)$。
耗时$3$ ms，排名$89.21\%$

Better

思路

更多的判环算法看这里Wiki。