LeetCode 30 days Challenge - Day 2
本系列将对LeetCode新推出的30天算法挑战进行总结记录，旨在记录学习成果、方便未来查阅，同时望为广大网友提供帮助。
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:
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1
Solution
题目要求分析：给定一个整数，将该整数每一位平方加和得到新的整数，重复直到整数为1或出现循环。
解法：
快慢指针：遇到出现循环情况的题目，可以考虑快慢指针的方法：
- 快指针：一次走2步（在本题中即进行两次运算）。
- 慢指针：一次走1步（进行一次运算）。
在若干次迭代后，如果存在循环，快慢指针会指向同一值，此时退出循环即可。
class Solution {
public:
int get_pow(int n) {
int tmp = 0;
while (n) {
tmp += (n % 10) * (n % 10);
n /= 10;
}
return tmp;
}
bool isHappy(int n) {
if (n == 1) return true;
int slow = n, fast = n;
do {
slow = get_pow(slow);
fast = get_pow(fast);
fast = get_pow(fast);
} while (slow != fast);
return (slow == 1);
}
};
传送门：Happy Number
2020/4 Karl
