169 lines
3.8 KiB
C++

// Source : https://oj.leetcode.com/problems/reverse-words-in-a-string/
// Author : Hao Chen, Siwei Xu
// Date : 2014-06-16
/**********************************************************************************
*
* Given an input string, reverse the string word by word.
*
* For example,
* Given s = "the sky is blue",
* return "blue is sky the".
*
*
* 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.
*
*
**********************************************************************************/
#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <string>
#include <vector>
#include <algorithm> // for std::reverse
using namespace std;
void reverseWords(string &s) {
bool wordStart = false;
vector<string> v;
const char *pHead =s.c_str();
const char *pStr, *pBegin, *pEnd;
for (pStr=pHead; *pStr!='\0'; pStr++) {
if (!isspace(*pStr) && wordStart == false){
wordStart = true;
pBegin = pStr;
continue;
}
if(isspace(*pStr) && wordStart==true){
wordStart=false;
pEnd = pStr;
v.insert(v.begin(), s.substr(pBegin-pHead, pEnd-pBegin) );
}
}
if (wordStart==true){
pEnd = pStr;
v.insert(v.begin(), s.substr(pBegin-pHead, pEnd-pBegin) );
}
if (v.size()>0){
s.clear();
char space=' ';
vector<string>::iterator it;
for (it=v.begin(); it!=v.end(); ++it) {
s = s + *it;
s.push_back(space);
}
s.erase(s.end()-1);
}else{
s.clear();
}
cout << "[" << s << "]" <<endl;
}
// inspired from <Programming Pearls> -- Handwaving
void reverseWords2(string &s) {
if (s.length() == 0) return;
string result = "";
if (s[s.length()-1] == ' ') {
int last = s.find_last_not_of(' ') + 1;
s.erase(last, s.length() - last);
}
int first = s.find_first_not_of(' ', 0);
while (first != string::npos) {
int wend = s.find(' ', first); // word end
if (wend == string::npos) wend = s.length();
string word = s.substr(first, wend - first);
reverse(word.begin(), word.end());
result += word;
first = s.find_first_not_of(' ', wend); // next word
if (first == string::npos) break;
result += ' ';
}
reverse(result.begin(), result.end());
s.swap(result);
}
// C solution in O(1) space
void reverse(char *b, char *e) {
for (--e; e - b > 0; b++, e--) {
char t = *b;
*b = *e;
*e = t;
}
}
void reverseWords(char *s) {
char *p = s, *ws = NULL, *last = s;
while (*p && *p == ' ') p++; // skip leading space
ws = p;
for ( ; *p; p++) {
while (*p && *p != ' ') p++; // find word end
reverse(ws, p);
strncpy(last, ws, p-ws);
last += (p-ws);
while (*p && *p == ' ') p++; // for next word
ws = p;
if (*p == '\0') break;
*last++ = ' ';
}
reverse(s, last);
*last = '\0';
}
void test() {
#define TEST(str) do { \
char* s = strdup(str); \
printf("\"%s\" => ", s); \
reverseWords(s); \
printf("\"%s\"\n\n", s); \
free(s); \
} while (0)
TEST(" the blue sky is blue ");
TEST(" ");
}
main()
{
string s;
reverseWords(s);
s=" ";
reverseWords(s);
s="1 ";
reverseWords(s);
s="love";
reverseWords(s);
s="i love cpp";
reverseWords(s);
test();
}