Leetcode 记录——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: 19 is a happy number

12 + 92 =
82
82 + 22 =
68
62 + 82 =
100

12 + 02 +
02 = 1
这个用很简单的数学方法，就可以实现。

bool isHappy(int n) {
int remainder ;
long int sum = 0;
int times = 0;
while (n != 1 && times < 100)//循环判断，这里有些不足，因为Leetcode的测试用例有的好变态，所以用的100...更正确的方法是用哈//希表，如果出现重复的sum，则不是happy number，哈希表则用是否超过最大整数来做终止判断。	do{
remainder = n % 10;
sum += remainder * remainder;
} while ((n = n / 10) > 0);
n = sum;
sum = 0;
times++;
}
if (n == 1)
return true;
else
return false;
}

之前 做了一道题也用到hash table, Fraction to recurring decimal把分数变换成小数,中等题，用最笨的方法没做出来，也是判断是否出现循环的余数。
Hash table，散列表，介于线性表和树之间的数据结构，适合 查找、插入、删除操作，不适合排序；是将关键字映射到单元。C++有模板，C语言编程好麻烦(⊙o⊙)…；对于重复的映射，有两种方法解决，一种是链表数组的方法（分离连接法），一种是重新分配的方法（开放定址法）；