Sum of Triangular Numbers

Description:Your task is to return the sum of Triangular Numbers up-to-and-including the nth Triangular Number.Triangular Number: "any of the series of numbers (1, 3, 6, 10, 15, etc.) obtained by continued summation of the natural numbers 1, 2, 3, 4, 5, etc."[01]02 [03]04 05 [06]07 08 09 [10]11 12 13 14 [15]16 17 18 19 20 [21]e.g. If 4 is given: 1 + 3 + 6 + 10 = 20.Triangular Numbers cannot be negative so return 0 if a negative number is given.

题目如上

自己的的渣渣解法

int sumTriangularNumbers(int n){ return num(n , 1 , 0);}//n层数  m层次起始   count递归统计int num(int n , int m , int count){  int i = 0 ;  if(count < n){    for(i = 0 ; i < count ; i++){      ++m;    }    return m + num(n , (m+1) , ++count);  }else{    return 0;  }}

别人优秀的解法

更好的规律 层次等于层次的累加

int sumTriangularNumbers(int n){  // triangular number  int t = 0;  // sum  int s = 0;  for (int i = 1; i < n+1; i++){    t = t + i;    s = s + t;  }  return s;}

以及

这个不知道推导过程 = =

int sumTriangularNumbers(int n){  if(n<=0){    return 0;  }  return n*(n+1)*(n+2)/6;}

后记

在同学的推荐下,进入了codewars真的能学到东西,也能锻炼思维.