满意答案

nanhrui57q

2014.01.10

采纳率：57%    等级：12

已帮助：8634人

#include

#include

int compear(void *, void *);

int delEle(int *, int, int);

int main(int argc, char *argv[])

{

int len = 10;

int arrs[10] = {4, 1, 5, 2, 7, 3, 9, 0, 6, 8};

int i;

qsort(arrs, 10, sizeof(int), compear);

for(i = 0; i < 10; i++)

printf("%d ", arrs[i]);

printf("\n");

len = delEle(arrs, 2, len);//delete 2

printf("\n");

len = delEle(arrs, 11, len);//delete 11 but doesn't exist

rdel(arrs, 2, len, 0);

return 1;

}

int compear(void *a, void *b)

{

int *ma = (int *)a, *mb = (int *)b;

if(*ma > *mb)

return 1;

if(*ma == *mb)

return 0;

if(*ma < *mb)

return -1;

}

int delEle(int *arr, int ele, int len)

{

int i, j, found = 0, mlen = len;

for(i = 0; i < len; i++){

if(arr[i] == ele){

for(j = i; j < len - 1; j++){

arr[j] = arr[j + 1];

}

found = 1;

mlen--;

}

}

if(found == 0)

printf("element %d doesn't exists!\n", ele);

for(i = 0; i < mlen; i++)

printf("%d ", arr[i]);

return mlen;//return the length of the arrays

}

//递归删除

int rdel(int *ar, int ele, int len, int flag)//flag为是否前面删除过元素的标记

{

int tmp;

if(len == 1)//数组结束

return *ar;

if(*ar == ele || flag == 1){//发现删除元素或前面删过元素 后面的元素全部前移一位

tmp = *ar;

*ar = rdel(ar + 1, ele, len - 1, 1);

return tmp;

}else{

rdel(ar + 1, ele, len - 1, 0);//前面未删 跳过

}

递归这个比较粗糙 没有边界检查，删最后一个可能有点问题 自己修改下好了

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