202. 快樂數 - 力扣(LeetCode)
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思路:
可以借用判斷鏈表是否有環的思想:
- ?定義快慢指針(兩個變量賦值就行)
- 快指針走兩次,慢指針走一次
- 快慢指針相遇,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;}