假定数组A[nSize]中有多个0元素，写一个函数，将A中所有非0元素一次移动数组A的前端A[i](0<=i<=nSize)，

//C
#include <stdio.h>
#define nSize 20
int fun(int A[])
{
int i,j,tmp;
for (i = 0; i < nSize-1; ++i)
{
if (A[i] == 0)
{
for (j = i+1; j < nSize; ++j)
{
if (A[j] != 0)
{
tmp = A[i];
A[i] = A[j];
A[j] = tmp;
break;
}
}
}
}
for (i = 0,tmp = 0; i < nSize; ++i)
if (A[i] != 0) tmp++;
return tmp;
}
void main()
{
int i,count;
int A[nSize]={0,1,2,3,4,0,6,7,8,9,0,11,12,13,14,0,16,17,18,0};
count = fun(A);
for (i = 0; i < nSize; ++i)printf("%d,", A[i]);
printf("\b\t\ni = %d\n", count);
}

写个java的，抛砖引玉

public int getCount(int[] A){
int count = 0;
for(int i=0; i<A.length; i++){
if(A[i] != 0){
int temp = A[0];
A[0] = A[i];