recursion - Java Square function -

would able explain these methods few comments line. squaring number. 1 using recursion, implementation , normal

 public static int sq(int n) {     int = 0 ;     int result = 0 ;      while(i < n){                  result = result + 2*i + 1 ;         = + 1 ;     }     return result ; }  public static int recsq(int n) {     if(n == 0){         return 0 ;     } else {         return recsq(n - 1) + 2*(n - 1) + 1 ;     } }  public static int implementsq(int n) {     int ;     int result = 0 ;      for(i = 0 ; < n ; i++){         result = result + 2*i + 1 ;            }     return result ; }  

• the first 1 multiplying number 2 n times using loop increase local variable i.
• the second 1 doing same using recursion. each step decreasing n , returning 0 final case. calls calling again same function different parameters exept parameter value 0, function return 0. recursion not simple thing think off, understand better try imagine codeflow.

example: recsq(2)

  4 recsq(2)   |<- 1  reqsq(1)+2*1+1       |<-0  reqsq(0)+2*0 + 1          |<-  0 

reqsq(2) called eval if , start evaluating return statement. first operation method call reqsq(n-1) n = 2 call reqsq(1).

reqsq(1) called eval if , start evaluating return statement. first operation method call reqsq(n-1) n = 1 call reqsq(0).

reqsq(0) called eval if, it's true cause n ==0 , return 0.

reqsq(1) call has finished evaluating reqsq(0) , can proceed calculating rest 0 + 2*(n-1) + 1 -> 0 + 0 + 1. return value 1.

reqsq(2) has finished evaluating reqsq(1) , can proceed calculating rest 1 + 2*(n-1) +1 -> 1 + 2 + 1. return value 4.

• the last 1 loop, practically same first 1 using fors instead of while loops. in loop declare in 1 line initialization condition, continue condition , increase operation. in case loop starting value 0, loop continue < n , @ end of loop "i++" called.