本文介绍了两个数组的交集 - Ç的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
由于某种原因，当
SIZEOF_A = 6
SIZEOF_B = 4
集合A = {0，1，2，3，4，5，6}
集合B = {1，10，11，12}
交集的所有扭曲（ 1048832 ）。
这是为什么？
的#include＆QUOT; usefunc.h＆QUOT;/ *＃定义SIZEOF_A 2
＃定义SIZEOF_B 4 * /的typedef枚举{
A，B，C，D，E，F，G，H，I，J，K，L，M，N，O，P，Q，R，S，T，U，V，W，X，Y， ž
}设置;无效dispSet（组号码[]，诠释size_numbers）{
INT I;
的printf（QUOT; [＆QUOT）;
对于（i = 0; I＆LT; size_numbers-1;我++）{
的printf（QUOT;％D，＆QUOT ;,号[I]）;
}
的printf（QUOT;％d个] QUOT ;,号码[size_numbers-1]）;
的printf（QUOT; \\ n＆QUOT;）;
}INT in_arr（INT A，集合B []，诠释B_size）{
INT RES = 0;
INT计数器;
对于（计数器= 0;反＆LT; B_size;反++）{
若（a == B〔计数器]）{
RES = 1;
打破;
}
}
返回水库;
}INT arr_in_arr（设定较小的[]，INT smaller_size，设置更大的[]，诠释bigger_size）{
中期业绩= 1;
INT计数器;
对于（计数器= 0;反＆LT; smaller_size;反++）{
如果（in_arr（小[计数器]，做大，bigger_size））{
继续;
}
其他{
RES = 0;
打破;
}
}
返回水库;
}INT size_c（设置ARR1 []，INT arr1_size，设置ARR2 []，诠释arr2_size）{
INT I;
INT newsize = 0;
对于（i = 0; I＆LT; arr1_size;我++）{
如果（！in_arr（ARR1 [I]，ARR2，arr2_size））{
newsize ++;
}
}
对于（i = 0; I＆LT; arr2_size;我++）newsize ++;
返回newsize;
}INT size_d（设置ARR1 []，INT arr1_size，设置ARR2 []，诠释arr2_size）{
INT I;
INT SIZEOF_D = 0;
对于（I = 0; I＆下; arr1_size;我++）如果（in_arr（ARR1 [I]，ARR2，arr2_size））SIZEOF_D ++;
返回SIZEOF_D;
}INT加入（设置ARR1 []，INT arr1_size，设置ARR2 []，INT arr2_size，设置ARR3 []，诠释arr3_size）{
INT I，J = 0;
对于（i = 0; I＆LT; arr1_size;我++）ARR3 [I] = ARR1 [I] 对于（i = 0; I＆LT; arr2_size;我++）{
如果（！in_arr（ARR2 [I]，ARR3，arr3_size））{
ARR3 [J + arr1_size] = ARR2 [I]
J ++;
}
}
}INT交集（设置ARR1 []，INT arr1_size，设置ARR2 []，INT arr2_size，设置ARR3 []，诠释arr3_size）{
INT I，J = 0;
对于（i = 0; I＆LT; arr1_size;我++）{
如果（in_arr（ARR1 [I]，ARR2，arr2_size））ARR3 [J] = ARR1 [I];
J ++;
}
}INT GETSET（ARR设置[]，INT大小）{
INT I;
的printf（QUOT;具有= 0开始，进入对应于一个字母一个数字\\ n＆QUOT;。）;
对于（i = 0; I＆LT;大小;我++）{
的printf（QUOT;民：＆QUOT;）;
改编[I] = GetInteger（）;
}
}诠释主（）{
的printf（QUOT; A＆QUOT的大小;）; INT SIZEOF_A = GetInteger（）;
的printf（QUOT; b的大小和QUOT;？）; INT SIZEOF_B = GetInteger（）;
设A [SIZEOF_A];的printf（QUOT; \\ NA：\\ n＆QUOT;）; GETSET（A，SIZEOF_A）;
集合B [SIZEOF_B]的printf（＆QUOT; B：\\ n＆QUOT;）; GETSET（B，SIZEOF_B）;
INT SIZEOF_C = size_c（A，SIZEOF_A，B，SIZEOF_B）;
INT SIZEOF_D = size_d（A，SIZEOF_A，B，SIZEOF_B）;
的printf（QUOT; d尺寸：％D＆QUOT ;, SIZEOF_D）;
设置C [SIZEOF_C];设置D [SIZEOF_D];
加入（A，SIZEOF_A，B，SIZEOF_B，C，SIZEOF_C）;
交集（A，SIZEOF_A，B，SIZEOF_B，D，SIZEOF_D）;
的printf（QUOT; \\ n有关集合\\ n＆QUOT;）;
的printf（QUOT; A：＆QUOT;）; dispSet（A，SIZEOF_A）;
的printf（＆QUOT; B：＆QUOT;）; dispSet（B，SIZEOF_B）;
的printf（QUOT; \\ n＆QUOT;）;的printf（QUOT;工会A，B：＆QUOT;）; dispSet（C，SIZEOF_C）;
的printf（QUOT;交点A，B：＆QUOT;）; dispSet（D，SIZEOF_D）;
的printf（QUOT;％S \\ n \\ n＆QUOT;，（arr_in_arr（A，SIZEOF_A，B，SIZEOF_B）== 1）＆QUOT; B包含A＆QUOT;：＆QUOT; B不包含A＆QUOT;）;
}
解决方案
在十字路口，你的if语句应该是：
如果（in_arr（ARR1 [I]，ARR2，arr2_size））{
ARR3 [J] = ARR1 [I]
J ++;
}
除非你找到一个匹配，不要增加学家
for some reason, when
SIZEOF_A = 6
SIZEOF_B = 4
set A = {0, 1, 2, 3, 4, 5, 6}
set B = {1, 10, 11, 12}
the intersection is all screwy (1048832).why is this?
#include "usefunc.h"
/* #define SIZEOF_A 2
#define SIZEOF_B 4 */
typedef enum {
a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
} set;
void dispSet(set numbers[], int size_numbers) {
int i;
printf("[ ");
for (i = 0; i < size_numbers-1; i++) {
printf("%d, ", numbers[i]);
}
printf("%d ]", numbers[size_numbers-1]);
printf("\n");
}
int in_arr(int A, set B[], int B_size) {
int res = 0;
int counter;
for (counter = 0; counter < B_size; counter++) {
if (A == B[counter]) {
res = 1;
break;
}
}
return res;
}
int arr_in_arr(set smaller[], int smaller_size, set bigger[], int bigger_size) {
int res = 1;
int counter;
for (counter = 0; counter < smaller_size; counter++) {
if (in_arr(smaller[counter], bigger, bigger_size)) {
continue;
}
else {
res = 0;
break;
}
}
return res;
}
int size_c(set arr1[], int arr1_size, set arr2[], int arr2_size) {
int i;
int newsize = 0;
for (i = 0; i < arr1_size; i++) {
if (!in_arr(arr1[i], arr2, arr2_size)) {
newsize++;
}
}
for (i = 0; i < arr2_size; i++) newsize++;
return newsize;
}
int size_d(set arr1[], int arr1_size, set arr2[], int arr2_size) {
int i;
int SIZEOF_D = 0;
for (i = 0; i < arr1_size; i++) if (in_arr(arr1[i], arr2, arr2_size)) SIZEOF_D++;
return SIZEOF_D;
}
int Join(set arr1[], int arr1_size, set arr2[], int arr2_size, set arr3[], int arr3_size) {
int i, j = 0;
for (i = 0; i < arr1_size; i++) arr3[i] = arr1[i];
for (i = 0; i < arr2_size; i++) {
if (!in_arr(arr2[i], arr3, arr3_size)) {
arr3[j+arr1_size] = arr2[i];
j++;
}
}
}
int Intersection(set arr1[], int arr1_size, set arr2[], int arr2_size, set arr3[], int arr3_size) {
int i, j = 0;
for (i = 0; i < arr1_size; i++) {
if (in_arr(arr1[i], arr2, arr2_size)) arr3[j] = arr1[i];
j++;
}
}
int getSet(set arr[], int size) {
int i;
printf("Starting with a=0, enter a number that corresponds to a letter.\n");
for (i = 0; i < size; i++) {
printf("NUM: ");
arr[i] = GetInteger();
}
}
int main () {
printf("size of a? "); int SIZEOF_A = GetInteger();
printf("size of b? "); int SIZEOF_B = GetInteger();
set A[SIZEOF_A]; printf("\nA:\n"); getSet(A, SIZEOF_A);
set B[SIZEOF_B]; printf("B:\n"); getSet(B, SIZEOF_B);
int SIZEOF_C = size_c(A, SIZEOF_A, B, SIZEOF_B);
int SIZEOF_D = size_d(A, SIZEOF_A, B, SIZEOF_B);
printf("size of D: %d", SIZEOF_D);
set C[SIZEOF_C]; set D[SIZEOF_D];
Join(A, SIZEOF_A, B, SIZEOF_B, C, SIZEOF_C);
Intersection(A, SIZEOF_A, B, SIZEOF_B, D, SIZEOF_D);
printf("\nFor the sets\n");
printf("A: "); dispSet(A, SIZEOF_A);
printf("B: "); dispSet(B, SIZEOF_B);
printf("\n"); printf("union A,B: "); dispSet(C, SIZEOF_C);
printf("intersection A,B: "); dispSet(D, SIZEOF_D);
printf("%s\n\n", (arr_in_arr(A, SIZEOF_A, B, SIZEOF_B) == 1)?"B contains A":"B does not contain A");
}
解决方案
In Intersection, your if statement should be:
if (in_arr(arr1[i], arr2, arr2_size)) {
arr3[j] = arr1[i];
j++;
}
Don't increment j unless you find a match.
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