letecode [202] - 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: 12 + 92 = 82 82 + 22 = 68 62 + 82 = 100 12 + 02 + 02 = 1
题目大意:
给定正整数,按照示例方式用它每个数的平方和替代它,若平方和等于1则该正整数为快乐数;若一直无限循环则非快乐数。
理 解:
对于正整数n,n%10从后往前遍历它每一个数,n = n/10删除已遍历的数,sum表示平方和,若sum=1,则n为快乐数;否则,n = sum重新计算平方和。
用集合set判断是否存在循环,每计算一个数,就保存在set中,若s.count(n)不等于0表明已经计算过该数,出现了循环,n为非快乐数。
代 码 C++:
class Solution { public: bool isHappy(int n) { int sum = 0; set<int> s; while(n!=1){ s.insert(n); while(n!=0){ sum += (n%10)*(n%10); n = n/10; } n = sum; if(s.count(n)!=0) return false; sum = 0; } return true; } };
运行结果:
执行用时 :4 ms, 在所有C++提交中击败了97.12%的用户
内存消耗 :8.5 MB, 在所有C++提交中击败了11.41%的用户