刷题-LeetCode151 Reverse Words in a String
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- Reverse Words in a String
Given an input string, reverse the string word by word.
Example 1:
Input: "the sky is blue"
Output: "blue is sky the"
Example 2:
Input: " hello world! "
Output: "world! hello"
Explanation: Your reversed string should not contain leading or trailing spaces.
Example 3:
Input: "a good example"
Output: "example good a"
Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.
Note:
- A word is defined as a sequence of non-space characters.
- Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
- You need to reduce multiple spaces between two words to a single space in the reversed string.
解法1 分词,然后用字符串拼接
class Solution {
public:
string reverseWords(string s) {
string tmp, res;
for(int i = 0; i < s.size(); ++i){
if(s[i] == ‘ ‘){
if(tmp.size() > 0){
res = tmp + " " + res;
tmp = "";
}
}else{
tmp += s[i];
}
}
res = tmp + " " + res;
int i = 0, j = res.size() - 1;
while(i < s.size() && res[i] == ‘ ‘)i++;
while(j >= 0 && res[j] == ‘ ‘)j--;
return res.substr(i, j-i+1);
}
};
解法2 字符串翻转。先将字符串整体翻转,然后将每个单词翻转。注意需要预处理除掉字符串首尾的空格,翻转结束后需要去除字符串中间多余的空格
class Solution {
public:
string reverseWords(string s) {
while(s.size() > 0 && s[0] == ‘ ‘)s.erase(0,1);
while(s.size() > 0 && s[s.size()-1] == ‘ ‘)s.erase(s.size()-1, 1);
reverse(s, 0, s.size()-1);
int pre = 0, cur = 0;
while(cur < s.size()){
if(s[cur] == ‘ ‘){
reverse(s, pre, cur-1);
cur += 1;
pre = cur;
}else{
cur += 1;
}
}
reverse(s, pre, cur-1);
int i = 1;
while(i < s.size()){
if(s[i] == ‘ ‘ && s[i-1] == ‘ ‘)s.erase(i,1);
else i++;
}
return s;
}
void reverse(string &s, int pre, int cur){
while(pre < cur){
char tmp = s[pre];
s[pre] = s[cur];
s[cur] = tmp;
pre++;
cur--;
}
}
};
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