leetcode 題號202 Happy Number

查看題目詳情可點擊此處。

題目

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:

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

解題思路

其實題目是想一個將數字的每一位平方的和作爲一個新數字，這個數字如果爲 1，那它就是 Happy Num，如果不是 1 將這個新數字計算每一位的平方之和，又以此爲一個新的數字，直到計算得到 1 或者計算的數字已經出現過就結束運算。

計算過的數字需要一個容器存儲，並且能夠快速查找，那 HashMap 再合適不過，最終就是計算每一位的平方和進行判斷。

代碼實現

class Solution {
    private static final int DIGIT_NUM = 10;
    private static final int[] DIGIT_SQUARE_NUMS = new int[]{0, 1, 4, 9, 16, 25, 36, 49, 64, 81};
    
    public boolean isHappy(int n) {
        int num = n;
        Map<Integer, Integer> computeNums = new HashMap<>();
        while (num != 1) {
            num = sumSquareDigits(num);
            if (computeNums.containsKey(num)) {
                return false;
            }

            computeNums.put(num, 1);
        }
        return true;
    }
    
    public static int sumSquareDigits(int num) {
        int sum = 0;
        while (num != 0) {
            sum += DIGIT_SQUARE_NUMS[num % DIGIT_NUM];
            num /= DIGIT_NUM;
        }

        return sum;
    }
}

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代碼詳情可點擊查看 我的 GitHub 倉庫