The number of N arrays is repeated n/2 + 1 times.

Source: Internet
Author: User

Problem:

An array with a length of N, one of which repeats n/2 + 1 times to find this number. solution: the direct idea is to traverse each element, compare it with other elements, equal to the counter sum ++ once until sum = n/2 + 1;
# Include <stdio. h> # Include <Stdlib. h> # Include <Assert. h> Int Fun ( Int Indium [], Int  Size) {assert (Indium ! = NULL & size> 1  ); Int I = 0 , J = 0  ;  For (; I <size- 1 ; I ++ ){  Int Sum = 1  ;  For (J = I + 1 ; J <size; j ++ ){  If (Indium [I] = indium [J]) sum ++;}  If (Size/ 2 + 1 = Sum) Return  Indium [I];}  Return - 1  ;}  Int  Main (){  Int Input [] = { 1 , 2 , 5 ,2 , 1 , 8 , 2 , 2 , 2 , 2  };  Int Ret = fun (input, 10  ); Printf (  "  Result = % d \ n  "  , RET ); Return   0  ;} 

The result is as follows:

 
[[Email protected] desktop] #./.OutResult=2[[Email protected] desktop] #

 

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