LeetCode 202:Happy Number
2015-05-08 22:12
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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
分析:
首先我们需要计算数字N每一位平方和,这个比较简单,代码如下:
然后就是函数的主体部分,需要注意的是,可能会出现循环的情况,如 :
2 、4、16、37、58、89、145、42、20、4、16、37。。。
所以我们需要用一集合将所有出现过的n和countN都记录下来,这样再次出现时能够及时的停止程序的执行避免出现死循环。Map是一个比较好的选择,因为查找元素的复杂度为O(1)。
主体部分代码如下:
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
分析:
首先我们需要计算数字N每一位平方和,这个比较简单,代码如下:
int countN(int n){ int tmp = n; int sum = 0; while (tmp > 0) { int result = tmp % 10; sum += result * result; tmp = tmp /10; } return sum; }
然后就是函数的主体部分,需要注意的是,可能会出现循环的情况,如 :
2 、4、16、37、58、89、145、42、20、4、16、37。。。
所以我们需要用一集合将所有出现过的n和countN都记录下来,这样再次出现时能够及时的停止程序的执行避免出现死循环。Map是一个比较好的选择,因为查找元素的复杂度为O(1)。
主体部分代码如下:
bool isHappy(int n){ map<int, int> int_map; if (n==1) { return true; } int sum = 0; while (n != 1) { int_map.insert(pair<int , int>(n, n)); sum = countN(n); if (sum ==1) { return true; } if(int_map.count(sum)) return false; n = sum; } return true; }
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