58. 最后一个单词的长度
- 58. 最后一个单词的长度
- 思路：反向遍历
- 时间：O(n)；空间：O(1)
class Solution {
public:
int lengthOfLastWord(string s) {
int id = s.size() - 1;
while(s[id] == ' '){
--id;
}
int ret = 0;
while(id >= 0 && s[id] != ' '){
ret++;
id--;
}
return ret;
}
};
1768. 交替合并字符串
- 1768. 交替合并字符串
- 思路：双指针，类似归并排序
- 时间：O(max(m, n))；空间：O(1)
class Solution {
public:
string mergeAlternately(string word1, string word2) {
string ret;
int p = 0, q = 0;
int w1_size = word1.size(), w2_size = word2.size();
while(p < w1_size && q < w2_size){
ret += word1[p++];
ret += word2[q++];
}
while(p < w1_size){
ret += word1[p++];
}
while(q < w2_size){
ret += word2[q++];
}
return ret;
}
};
