Reverse Words in a string

Posted 2020-08-09 amazingzoe

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.

 

Analyse: first reverse the entire string, then have wordBegin as the index for the starting position of the word, find the left and right index of a word, reverse from left to right, copy left to right to wordBegin, set ‘ ‘ to the next of that word, set correct index for wordBegin.

Be aware of the empty string, the length of word equals to 1. 

Runtime: 9ms

 1 class Solution {
 2 public:
 3     void reverseWords(string &s) {
 4         if (s.empty()) return;
 5         reverse(s.begin(), s.end());
 6         
 7         int left = 0, right = 0, wordBegin = 0;
 8         int index = 0;
 9         for (; index < s.size(); index++) {
10             if (isspace(s[index])) continue;
11             if (index == 0 || isspace(s[index - 1]))
12                 left = index;
13             if (index == s.size() - 1 || isspace(s[index + 1])) right = index;
14             else continue;
15             
16             reverse(s.begin() + left, s.begin() + right + 1);
17             // leading 0s before the word
18             if (wordBegin < left) {
19                 while (left <= right) {
20                     s[wordBegin++] = s[left++];
21                 }
22                 s[wordBegin] = ‘ ‘;
23                 wordBegin++;
24             } else {
25                 wordBegin = right + 2;
26             }
27         }
28         wordBegin = wordBegin ? wordBegin : 1;
29         s.erase(s.begin() + wordBegin - 1, s.end());
30     }
31 };