题目描述
给你两个字符串 word1
和 word2
。请你从 word1
开始,通过交替添加字母来合并字符串。如果一个字符串比另一个字符串长,就将多出来的字母追加到合并后字符串的末尾。
返回 合并后的字符串 。
https://leetcode.cn/problems/merge-strings-alternately/description/
示例
示例 1:
输入:word1 = "abc", word2 = "pqr"
输出:"apbqcr"
解释:字符串合并情况如下所示:
word1: a b c
word2: p q r
合并后: a p b q c r
示例 2:
输入:word1 = "ab", word2 = "pqrs"
输出:"apbqrs"
解释:注意,word2 比 word1 长,"rs" 需要追加到合并后字符串的末尾。
word1: a b
word2: p q r s
合并后: a p b q r s
示例 3:
输入:word1 = "abcd", word2 = "pq"
输出:"apbqcd"
解释:注意,word1 比 word2 长,"cd" 需要追加到合并后字符串的末尾。
word1: a b c d
word2: p q
合并后: a p b q c d
提示:文章来源:https://www.toymoban.com/news/detail-839222.html
1 <= word1.length, word2.length <= 100
-
word1
和word2
由小写英文字母组成
解题总结
实现方式1是我自己的写法,写完之后感觉有点冗余,但又不知如何改进,实现方式2则是借鉴了别人的题解,显然,方式2的代码更加简洁优雅文章来源地址https://www.toymoban.com/news/detail-839222.html
代码实现
- 实现方式1
char * mergeAlternately(char * word1, char * word2){
int length1 = strlen(word1);
int length2 = strlen(word2);
char* res = (char*)malloc(sizeof(char) * (length1 + length2 + 1));
int i = 0;
int j = 0;
while(word1[i] != 0 && word2[i] != 0)
{
res[j] = word1[i];
j++;
res[j] = word2[i];
i++;
j++;
}
if(word1[i] == 0)
{
while(word2[i] != 0)
{
res[j] = word2[i];
i++;
j++;
}
}
else
{
while(word1[i] != 0)
{
res[j] = word1[i];
i++;
j++;
}
}
res[j] = 0;
return res;
}
- 实现方式2
char * mergeAlternately(char * word1, char * word2) {
int length1 = strlen(word1);
int length2 = strlen(word2);
char* res = malloc(sizeof(char) * (length1 + length2 + 1));
int i = 0;
int j = 0;
while (i < length1 || i < length2)
{
if (i < strlen(word1))
{
res[j++] = word1[i];
}
if (i < strlen(word2))
{
res[j++] = word2[i];
}
i++;
}
res[j] = 0;
return res;
}
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