Happy Number - LeetCode
examination questions
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
Please use the following function to solve the problem：
bool isHappy(int n) {
}
解题代码
#include <stdio.h> bool isHappy(int n) { int arr[5000]; int i = 0; int sum = 0; int temp1,temp2; while (true){ do{ temp1 = n / 10; temp2 = n % 10; n = temp1; sum = sum + temp2 * temp2; } while (temp1 > 0); if (sum != 1){ arr[i] = sum; n = sum; i++; } else{ return true; } for (int j = 1; j < i; j++){ if (sum == arr[j-1]){ return false; } } sum = 0; } } int main(){ int n; scanf("%d",&n); if (isHappy(n)){ printf("true\n"); } else{ printf("false\n"); } return 0; }
LeetCode判断结果:
基本算法思想
把一个数拆分成单个数, 然后进行平方和, 对和进行比较, 如果满足和为1, 就是快乐数, 如果进入循环, 那么就不是快乐数.
代码注释分析
#include <stdio.h> bool isHappy(int n) { int arr[5000]; //用于存储和不为1的值,仅仅是猜想有这么多个值 int i = 0; //记录和的个数 int sum = 0; //和值 int temp1, temp2; //temp1为值, temp2为余数 while (true){ do{ temp1 = n / 10; temp2 = n % 10; n = temp1; sum = sum + temp2 * temp2; //和 } while (temp1 > 0); //单个分解完成 if (sum != 1){ //如果不为1,就把该和值赋值给arr arr[i] = sum; n = sum; i++; } else{ return true; } for (int j = 1; j < i; j++){ //对和值进行循环匹配,如果和arr中的值相同,就退出,因为出现了循环 if (sum == arr[j - 1]){ return false; } } sum = 0;//第二次求和,必须先清零 } } int main(){ int n; scanf("%d", &n); if (isHappy(n)){ printf("true\n"); } else{ printf("false\n"); } return 0; }
优化
经过资料查询, 对于快乐数, 有以下规律:
不是快乐数的数称为不快乐数（unhappy number），所有不快乐数的数位平方和计算，最後都会进入 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4 的循环中。
优化代码
#include <stdio.h> bool isHappy(int n) { int sum = 0; int temp1, temp2; while (true){ do{ temp1 = n / 10; temp2 = n % 10; n = temp1; sum = sum + temp2 * temp2; } while (temp1 > 0); if (sum == 1){ return true; } else{ if (sum == 4 || sum == 20 || sum == 16 || sum == 37 || sum == 42 || sum == 58 || sum == 89 || sum == 145){ return false; } } n = sum; sum = 0; } } int main(){ int n; scanf("%d", &n); if (isHappy(n)){ printf("true\n"); } else{ printf("false\n"); } return 0; }
LeetCode判断结果:
