A solution in C/C++:
void swap(char* str, int i, int j){
char t = str[i];
str[i] = str[j];
str[j] = t;
}
void reverse_string(char* str, int length){
for(int i=0; i<length/2; i++){
swap(str, i, length-i-1);
}
}
void reverse_words(char* str){
int l = strlen(str);
//Reverse string
reverse_string(str,strlen(str));
int p=0;
//Find word boundaries and reverse word by word
for(int i=0; i<l; i++){
if(str[i] == ' '){
reverse_string(&str[p], i-p);
p=i+1;
}
}
//Finally reverse the last word.
reverse_string(&str[p], l-p);
}
This should be O(n) in time and O(1) in space.
Edit: Cleaned it up a bit.
The first pass over the string is obviously O(n/2) = O(n). The second pass is O(n + combined length of all words / 2) = O(n + n/2) = O(n), which makes this an O(n) algorithm.
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