- 题目: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 。 - 示例:
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
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
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
class Solution(object):
def mergeAlternately(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: str
"""
longword = ""
max_x = max(len(word1), len(word2))
word1 = word1.ljust(max_x)
word2 = word2.ljust(max_x)
longword = list(longword)
word1 = list(word1)
word2 = list(word2)
p = 0
for i in range(max_x):
longword.append(word1[i])
longword.append(word2[i])
longword = "".join(longword)
longword = re.sub(r"\s+", "", longword)
return longword
- 代码有点长,方法可以更精简。
- 将字符串转化为列表,采用append方法添加字符。
- 问题:在使用replace方法替换空格时,不知道为什么输出的还是有空格。
- 另一种解法:
class Solution(object):
def mergeAlternately(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: str
"""
max_x = max(len(word1), len(word2))
word1 = word1.ljust(max_x)
word2 = word2.ljust(max_x)
long_word = ""
for i in range(max_x):
long_word += word1[i]
long_word += word2[i]
long_word = re.sub(r"\s+", "", long_word)
return long_word
- 代码还是有点长,方法可以更精简。
- 直接使用str+来合并字符。