查看题目详情可点击此处。
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:
Input: 19
Output: true
Explanation:
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1
解题思路
其实题目是想一个将数字的每一位平方的和作为一个新数字，这个数字如果为 1，那它就是 Happy Num，如果不是 1 将这个新数字计算每一位的平方之和，又以此为一个新的数字，直到计算得到 1 或者计算的数字已经出现过就结束运算。
计算过的数字需要一个容器存储，并且能够快速查找，那 HashMap 再合适不过，最终就是计算每一位的平方和进行判断。
代码实现
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;
}
}
代码详情可点击查看 我的 GitHub 仓库
