1768. Merge Strings Alternately (使用C語言解題)

1768. Merge Strings Alternately

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Hint

You are given two strings word1 and word2. Merge the strings by adding letters in alternating order, starting with word1. If a string is longer than the other, append the additional letters onto the end of the merged string.

Return the merged string.

 

Example 1:

Input: word1 = "abc", word2 = "pqr"
Output: "apbqcr"
Explanation: The merged string will be merged as so:
word1:  a   b   c
word2:    p   q   r
merged: a p b q c r

Example 2:

Input: word1 = "ab", word2 = "pqrs"
Output: "apbqrs"
Explanation: Notice that as word2 is longer, "rs" is appended to the end.
word1:  a   b 
word2:    p   q   r   s
merged: a p b q   r   s

Example 3:

Input: word1 = "abcd", word2 = "pq"
Output: "apbqcd"
Explanation: Notice that as word1 is longer, "cd" is appended to the end.
word1:  a   b   c   d
word2:    p   q 
merged: a p b q c   d

 

Constraints:

• 1 <= word1.length, word2.length <= 100
• word1 and word2 consist of lowercase English letters.

 

google 翻譯：

給定兩個字串 word1 和 word2。透過以交替順序添加字母來合併字串，從 word1 開始。
如果一個字串比另一個字串長，請將附加字母附加到合併字串的末尾。

想法一：

使用一個 i 變數取 2 的餘數，
等於 0 時抓 word2 (如果 word2 已抓完了，就改抓 word1)，
等於 1 時抓 word1 (如果 word1 已抓完了，就改抓 word2)。

寫法一：

char * mergeAlternately(char * word1, char * word2){
    int len1= (word1 ?: strlen(word1));
    int len2 = (word2 ?: strlen(word2));
    int len = len1 + len2;
    char *str = malloc(len + 4);
    char *p1 = word1, *p2 = word2;
    int i = 0;
    while(*p1 || *p2) {
        if(i % 2) {
            if(*p2) {
                *(str+i) = *p2;
                p2++;
            } else {
                *(str+i) = *p1;
                p1++;
            }
        } else {
            if(*p1) {
                *(str+i) = *p1;
                p1++;
            } else {
                *(str+i) = *p2;
                p2++;
            }
        }
        i++;
    }
    *(str+i) = '\0';
    return str;
}

結果一：

速度尚可接受，但記憶體差別人太多了，看來有更簡潔的寫法

想法二：

參考了大神的寫法，發現跟本不需要 i 變數，跑一次迴圈直接看兩個 word 即可，大神真是厲害。

寫法二：

char * mergeAlternately(char * word1, char * word2){
    int len1= (word1 ? strlen(word1) : 0);
    int len2 = (word2 ? strlen(word2) : 0);
    char *str = malloc(len1 + len2 + 4);
    int i = 0, j = 0;
    while(i<len1 || i<len2) {
        if(i<len1) {
            *(str+j) = *(word1+i);
            j++;
        }
        if(i<len2) {
            *(str+j) = *(word2+i);
            j++;
        }
        i++;
    }
    *(str+j) = '\0';
    return str;
}

結果二：

結果已經很接近完美了

想法三：

再看一下其他大神有什麼想法可以參考的。
最後決定不使用 i、j 變數，改用 w1、w2 指標判斷字元，看起來更簡潔些。
再加上 isalnum 的錯誤判斷，就更沒有缺點了。

寫法三：

char * mergeAlternately(char * word1, char * word2){
    int len1= (word1 ? strlen(word1) : 0);
    int len2 = (word2 ? strlen(word2) : 0);
    char *str = malloc(len1 + len2 + 4);
    char *w1 = word1, *w2 = word2, *s = str;
    while(*w1 || *w2) {
        if(*w1) {
            if(isalnum(*w1)) {
                *s++ = *w1;
            }
            w1++;
        }
        if(*w2) {
            if(isalnum(*w2)) {
                *s++ = *w2;
            }
            w2++;
        }
    }
   *s = '\0';
    return str;
}

結果三：

速度和記憶體都還令人滿意~