202. 快乐数 - 力扣(LeetCode)
思路:
可以借用判断链表是否有环的思想:
- 定义快慢指针(两个变量赋值就行)
- 快指针走两次,慢指针走一次
- 快慢指针相遇,看是不是等于一
public int bitSum(int n){int sum = 0;while(n != 0){int m = n % 10;sum += m * m;n /= 10;}return sum;}public boolean isHappy(int n) {int fast = bitSum(n);int slow = n;while(slow != fast){slow = bitSum(slow);fast = bitSum(bitSum(fast));}return slow == 1;}