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关键是当一个数不是happy number是，怎样退出循环
自己的挫方法：使用set保存之前计算的所有数
class Solution {public: bool isHappy(int n) { if(n <= 0) return false; if(n == 1) return true; set prev; prev.insert(n); while(n != 1){ int sum = 0; while(n){ int digit = n % 10; sum += digit * digit; n /= 10; } n = sum; if(prev.count(n)) return false; prev.insert(n); } return true; }}; discuss里面的方法1： Using fact all numbers in [2, 6] are not happy (and all not happy numbers end on a cycle that hits this interval)
class Solution {public: bool isHappy(int n) { if(n <= 0) return false; while(n > 6){ int next = 0; while(n){ int tmp = n % 10; next += tmp * tmp; n /= 10; } n = next; } return n == 1; }}; discuss里面的方法2：Use the same way as checking cycles in a linked list
class Solution {public: bool isHappy(int n) { if(n <= 0) return false; int A = nextNum(n); //A指向第一个 int B = nextNum(A); //B指向第二个 while(A != 1 && A != B){ A = nextNum(A); //A每次向前走一步 B = nextNum(B); //B每次向前走两步 B = nextNum(B); } return A == 1; } int nextNum(int n){ int next = 0; while(n){ int tmp = n % 10; next += tmp * tmp; n /= 10; } return next; }}; 转载地址：http://llspi.baihongyu.com/