strStr

strStr.png

===================== 解題思路 =====================

基本的 nesting for loop 檢查從 source 的每一個字符為起點 能否與 target 成功比對, 第二層的 for loop 容易忘記 source 的起點是 i + j 而 target 永遠是 j 為起點 ( 從頭開始比 ), 另外就是對 source or target 可能是 nullptr 的情況在開頭先檢查一次, 另外也可以用 memcpy 來取得 substr 進行 strcmp 的比對, 但是基本上沒有太大的差別

===================== C++ code ====================

//first solution
 
 
class Solution {
 
 
public:
 
 
/**
 * Returns a index to the first occurrence of target in source,
 * or -1  if target is not part of source.
 * @param source string to be scanned.
 * @param target string containing the sequence of characters to match.
 */
int strStr(const char *source, const char *target) {
    if(source == nullptr || target == nullptr) return -1;
    int n1 = strlen(source), n2 = strlen(target);
    int i, j;
    for(i = 0; i <= n1 - n2 + 1; i++)
    {
        for(j = 0; j < n2; j++)
        {
            if(source[i+j] != target[j]) break;
        }
        if(j == n2) return i;
    }
    return -1;
}
 
 
};
 
 
//second solution (use memcpy to get substr)
 
 
class Solution {
 
 
public:
 
 
/**
 * Returns a index to the first occurrence of target in source,
 * or -1  if target is not part of source.
 * @param source string to be scanned.
 * @param target string containing the sequence of characters to match.
 */
int strStr(const char *source, const char *target) {
    if(source == NULL || target == NULL) return -1;
    int len_A = strlen(source);
    int len_B = strlen(target);
    for(int i = 0; i <= len_A - len_B; i++)
    {
        char tmp[len_B];
        memcpy( tmp, &source[i], len_B );
        tmp[len_B] = '\0';
        if(strcmp(tmp , target) == 0) return i;
    }
    return -1;
}
 
 
};