C-2数组，检查其中是否为子数组，并返回索引

Question

我有一个任务是编写一个函数，它将获得2个数组及其大小：

int contain(int big[], int size_b, int small[], int size_s) 

程序应该检查小数组是否是大数组的子数组，如果是，则返回小数组中第一个数字的索引号，否则返回-1。

示例:2 4 61 5 8 5 56 89 3 -2

5 56 89 3 -2

函数应返回5。

9 5 12 7 8 -2 4 32 900 13

9 5 12 8 7

函数应返回-1。

Answer 1

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int contain(int big[], int size_b, int small[], int size_s) {
    int contains, k;
    for(int i = 0; i < size_b; i++){
        // assume that the big array contains the small array
        contains = 1;
        // check if the element at index i in the big array is the same as 
        // the first element in the small array
       if(big[i] == small[0]){
           // if yes, then we start form k = 1, because we already know that
           // the first element in the small array is the same as the element 
           // at index i in the big array
           k = 1;
           // we start to check if the next elements in the big array
           // are the same as the elements in the small array
           // (we start from i+1 position because we already know that the element 
           // at the position i is the same as the first element in the small array)
           for(int j = i + 1; j < size_b; j++){
               // range for k must be from 1 to size_s-1
               // if we reached the end of the small array or if the small 
               // array contains only one element then break the for loop
               if(k >= size_s - 1) {
                   break;
               }
               // if the element at the position j in the big array is different
               // from the element at the position k in the small array then we
               // flag that we did not find the sequence we were looking for (contains=0)
               if(big[j] != small[k]){
                   contains = 0;
                   break;
               }
               // increment k because we want the next element in the small array
               k++;
           }
           // if contains flag is not 0 that means we found the sequence we were looking
           // for and that sequence starts from index i in the big array
           if(contains) {
               return i;
           }
       }
    }
    // if the sequence we were looking for was not found
    // then -1 is returned
    return -1;
}

以下是不带注释的代码

int contain(int big[], int size_b, int small[], int size_s) {
    int contains, k;
    for(int i = 0; i < size_b; i++){
        contains = 1;
       if(big[i] == small[0]){
           k = 1;
           for(int j = i + 1; j < size_b; j++){
               if(k >= size_s - 1) {
                   break;
               }
               if(big[j] != small[k]){
                   contains = 0;
                   break;
               }
               k++;
           }
           if(contains) {
               return i;
           }
       }
    }
    return -1;
}