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成績:將兩個已排序數組歸並成一個排序數組
這裡先不斟酌年夜數據量的情形(在數據量很年夜時不知年夜家有甚麼好的思緒或辦法?),只做簡略數組的處置。
簡略代碼以下:
解釋:之所以把merge函數界說成前往數組長度,是由於後續會有反復數據歸並功效的merge版本,斟酌到接口分歧性。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int merge(int* ar1, int len1, int* ar2, int len2, int** rtn)
/*++
DeScription:
This routine merge two sorted arrays into one sorted array,
the same values in different arrays will be keeped.
Arguments:
ar1 - The first sorted array to be merged
len1 - The num of items in ar1
ar2 - The second sorted array to be merged
len2 - The num of items in ar2
rtn - The caller proviced pointer to get the result array,
memory allocated for rtn should be free by the caller.
Return Value:
The num of items in the merge array
--*/
{
int i=0,j=0,k=0;
int m=0;
int* res = NULL;
if (ar1 == NULL || ar2 == NULL || rtn == NULL) {
return 0;
}
*rtn = (int *)malloc((len1+len2)*sizeof(int));
if(*rtn == NULL) {
return 0;
}
memset(*rtn, 0, (len1+len2)*sizeof(int));
res = (int*)*rtn;
while(i<len1 && j<len2) {
if (ar1[i]<=ar2[j]) {
res[k++] = ar1[i++];
} else {
res[k++] = ar2[j++];
}
}
while(i<len1) {
res[k++] = ar1[i++];
}
while(j<len2) {
res[k++] = ar2[j++];
}
return len1+len2;
}
int merge_test()
{
int a1[] = {0,1,2,5,8,19,34,43,52};
int a2[] = {1,4,5,12,17,33,42,51,53,65,76};
int len1 = sizeof(a1)/sizeof(int);
int len2 = sizeof(a2)/sizeof(int);
int i = 0, len = 0;
int* a3 = NULL;
int* ptr = NULL;
len = merge(a1, len1, a2, len2, &a3);
if (a3 == NULL) {
printf("a3==NULL\n");
return 1;
}
ptr = a3;
while(i<len) {
printf("a3[%3d]---->%8d\n", i++, *ptr++);
}
if (a3 != NULL) {
free(a3);
}
return 0;
}
int main(int argc, char* argv[])
{
merge_test();
return 0;
}