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
- 1^2 + 9^2 = 82
- 8^2 + 2^2 = 68
- 6^2 + 8^2 = 100
- 1^2 + 0^2 + 0^2 = 1
思路:给定一个输入后,有两种情况:是或者不是happy number。但是若不是的话,则判断过程会陷入无限循环。
这道题关键就是如何判断处于无限循环中。
这里有一个突破点,就是在判断happy number过程中,如果是无限循环,则循环中会有结果是重复出现的。
思路:设置一个slow变量,设置一个fast变量。然后我们定义一个函数get_sqsum(int n)来计算一个数n的各位数的平方和。
之后,在开始时slow和fast都等于n,然后在程序中我们进行循环,令slow = get_sqsum(slow)。 fast同理,但在每次循环中都进行两次。这相当于两个人在跑道上跑步,fast跑的速度是slow的两倍。因为跑道是圆的(loop),则fast必定会在某个时刻再次遇到slow,这时相当于已经领先了1整圈。
使用do while,循环结束的条件是判断slow是否等于fast。
这里可以看到,不管是否是happy number,slow和fast总有相等的时候。而如果是happy number,则循环结束时两个变量都应该等于1,否则不是。
1 class Solution { 2 public: 3 int get_sqsum(int n) 4 { 5 int res = 0; 6 while (n) 7 { 8 int tem = n % 10; 9 res += tem * tem; 10 n /= 10; 11 } 12 return res; 13 } 14 bool isHappy(int n) { 15 int slow, fast; 16 slow = fast = n; 17 do{ 18 slow = get_sqsum(slow); 19 fast = get_sqsum(fast); 20 fast = get_sqsum(fast); 21 } while (slow != fast); 22 return slow == 1; 23 } 24 };