Given an input string, reverse the string word by word.

For example,
Given s = "the sky is blue",
return "blue is sky the".

Update (2015-02-12):
For C programmers: Try to solve it in-place in O(1) space.

click to show clarification.

Clarification:

  • What constitutes a word?
    A sequence of non-space characters constitutes a word.
  • Could the input string contain leading or trailing spaces?
    Yes. However, your reversed string should not contain leading or trailing spaces.
  • How about multiple spaces between two words?
    Reduce them to a single space in the reversed string.

=====================

注意,

1,被空格包围的是单词

2,输入字符串可以以空格开头或结尾,但是结果中的字符不能以空格开头或结尾

3,输出字符串中单词间的空格是一个,不能重复出现空格.

思路:

对输入字符串进行去重空格操作,

对字符串中的每个单词进行反转

对整个字符串进行反转

====

code

class Solution {
public:
void help_reverse(string &s,int start,int end){
while(start<end){///经典的反转字符串方法
swap(s[start++],s[end--]);
}
}
string removeDuplicateSpace(string s){
string res;
int b = ;
for(;b<(int)s.size();b++){
if(s[b]!= ' '){
break;
}
}///
int e = s.size() - ;
for(;e>=;e--){
if(s[e]!=' '){
break;
}
} bool is_space = false;
for(int i = b;i<=e;i++){
if(s[i] == ' '){
if(!is_space) is_space = true;
else continue;
}else{
is_space = false;
}
res.push_back(s[i]);
}
return res;
}
void reverseWords(string &s) {
if(s.empty()) return;
s = removeDuplicateSpace(s);
int start = ;
for(size_t i = ;i<s.size();i++){
if(s[i]!=' '){
start = i;
}else{
continue;
}
size_t j = i;
while(j<s.size() && s[j]!=' '){
j++;
}
j--;
help_reverse(s,start,j);
i = j++;
}
help_reverse(s,,(int)s.size()-);
}
};

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