[英]finding if an array is a sub-sequence of another array
讓我們假設我有這兩個arrays
, int array[15] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
和int s_array[15] = {1,2,3,4,5}
*請注意,我知道s_array[]
設置為 15 但只有 5 個元素
在我的代碼的以下部分中,我想檢查s_array[]
是否是array[]
] 的子序列,這意味着存儲在s_array[]
中的任何值也存儲在array[]
中,並且在同一個命令。 這意味着{3,5,4}
不是子序列而{3,4,5}
是。
這是代碼的一部分(我認為問題所在)。 j
是s_array[]
中的數量。 整個代碼將在下面顯示
for( i = 0; i < 15; i++ ){
if( s_array[0] == array[i] ){ /*if the first element of the s_array is not
available then it is not a sub-sequence and
there is no need to check the rest */
for( Bcount = 1; Bcount < ( j - 1 ); Bcount++ ){
if( s_array[Bcount] == array[Bcount + i])
counter++;
else{
printf( "B is not a sub-sequence of A." );
break;
}
}
if( j == counter ){
printf( "B is a sub-sequence of A." );
break;
}
}
else
continue;
}
感興趣的人的完整代碼或者它是否有助於更好地理解
#include <stdio.h>
int main(){
int array[15], s_array[15], i = 0, j = 0, Bcount = 1, counter = 1;
printf( "Please enter a sequence A: " );
while(i < 15 && scanf("%d", &array[i]) == 1) {
//input for array[], will always have 15 elements
i++;
if( i == 15){
printf( "Please enter a sequence B: " );
while(j < 15 && scanf("%d", &s_array[j]) == 1) {
//input for s_array, will have somewhere between 1 and 15 elements
j++;
}
for( i = 0; i < 15; i++ ){
if( s_array[0] == array[i] ){
for( Bcount = 1; Bcount < ( j - 1 ); Bcount++ ){
if( s_array[Bcount] == array[Bcount + i])
counter++;
else{
printf( "B is not a sub-sequence of A." );
break;
}
}
if( j == counter ){
printf( "B is a sub-sequence of A." );
break;
}
}
else
continue;
}
return 5;
}
}
return 0;
}
您可以遍歷array
並為s_array
設置索引計數器。 如果僅array[i] == s_array[j]
,則增加該計數器,否則重置。
如果滿足以下條件,則可以更早地結束循環:
代碼示例:
#include <stdio.h>
#include<stdbool.h>
int main(void)
{
int array[15] = {1, 2, 3, 4, 6, 7, 6, 9, 0, 1, 2, 3, 4, 5, 6};
int s_array[5] = {2, 3, 4, 5, 6};
const int size_array = sizeof(array) / sizeof(array[0]);
const int size_s_array = sizeof(s_array) / sizeof(s_array[0]);
bool isFound = false;
if (size_s_array > size_array)
{
printf("Sub array is larger than the array\n");
}
else
{
int j = 0;
for (int i = 0; (i < size_array) && (j != size_s_array) ; i++)
{
if (array[i] == s_array[j])
{
j++;
}
else
{
j = 0;
if (i > size_array - size_s_array) // You can also put this in for loop's test condition, but I find this more readable.
{
break;
}
}
}
isFound = (j == size_s_array) ? true : false;
}
printf("%s\n", isFound ? "Found" : "Not found");
return 0;
}
不同s_array的測試用例:
array[15] = {1, 2, 3, 4, 6, 7, 6, 9, 0, 1, 2, 3, 4, 5, 6};
int s_array[5] = {2, 3, 4, 5, 6};
Found
int s_array[1] = {8};
Not found
int s_array[2] = {1, 2};
Found
int s_array[15] = {1, 2, 3, 4, 6, 7, 6, 9, 0, 1, 2, 3, 4, 5, 6};
Found
int s_array[16] = {0, 1, 2, 3, 4, 6, 7, 6, 9, 0, 1, 2, 3, 4, 5, 6};
Sub array is larger than the array
Not found
int s_array[2] = {0, 2};
Not found
謝謝#vitruvius。 我實現了他的邏輯並修改了部分代碼。 這里是。
#include<bits/stdc++.h>
using namespace std;
bool issub(int a[],int b[],int n,int m){
int j=0;
for(int i=0;i<n && j!=m;i++){
if(a[i]==b[j])
{
j++;
}
}
bool x=(j==m)?1:0;
return x;
}
int main(){
int n,m;
cin>>n>>m;
int a[n],b[m];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<m;i++){
cin>>b[i];
}
if(issub(a,b,n,m)){
cout<<"YES"<<endl;
}
else{
cout<<"NO"<<endl;
}
return 0;
}
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