Leecode on Two Sum of program, some places don't understand the
I guess this public int [] TwoSum (int [] Numbers, int target) statement is to define the array and the target,
But I don't understand public int usage, as well as the return of new int [] {I, j}; And return the new int [2]; What do you mean?

Public int [] TwoSum (int [] Numbers, int target)
{
For (int I=0; I & lt; Numbers. Length; I++)
{
For (int j=0; J & lt; Numbers. Length; J++)
{
If (I==j)
continue;

If (Numbers [I] + Numbers [j]==target)
Return new int [] {I, j};
}
}

Return new int [2];
}

CodePudding user response:

The other post I write my own:
#include
Int main ()
{
Int num [10]={,9,17,21,35,99,333,128,101,121, 2};
int i,j;
for(i=0; I<=9; I++)
{
for(j=0; J<=9; J++)
{
If (num + num [I] [j]==120)
{
Printf (" % d % d ", I, j);
I=100;
}
}
}
}

CodePudding user response:

 public int [] 

This is a C + + syntax, C language without public key;

Int [] such as the return value type is not common, common is int * return pointer, from the point of the return value of function is such, returns the pointer (address)

 # include & lt; Stdio. H> 
Int main () 
{
Int num [10]={,9,17,21,35,99,333,128,101,121, 2}; 

int i,j; 
for(i=0; I<=9; I++) 
{
for(j=0; J<=9; J++) 
{
If (I.=j & amp; & Num + num [I] [j]==120) 
{
Printf (" % d % d ", I, j); 
//I=100;//this assignment will lead to an array, didn't end because the second cycle could, if (num [I] here will appear an array; 
break; 
} 
} 
If (j & lt;=9)//continue to jump out of the loop 
break; 
} 
} 

For your reference ~

A bit of a problem to repair the code, as above;

CodePudding user response:

Tell it like it is, in this section, this problem is to zhao old wet, you @ zhao old wet,