LeetCode - Reverse Words in a String

Reverse Words in a String

2014.3.18 03:09

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

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

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.

Solution:

　　Reverse the whole string first, then reverse every single word. Redundant spaces must be skipped.

　　Total time complexity is O(n). Space complexity is O(1).

Accepted code:

 1 // 1TLE, 1AC, using another char[] is unnecessary. Don't miss '++i' or '++j'.

 2 class Solution {

 3 public:

 4     void reverseWords(string &s) {

 5         int i, j;

 6         int len;

 7         int offset;

 8         

 9         // remove trailing spaces

10         while (s.length() > 0 && s[s.length() - 1] == ' ') {

11             s.pop_back();

12         }

13         len = (int)s.length();

14         if (len == 0) {

15             return;

16         }

17         

18         // remove leading spaces

19         i = 0;

20         while (i < len && s[i] == ' ') {

21             ++i;

22         }

23         s = s.substr(i, len - i);

24         len = (int)s.length();

25         

26         // reverse the whole string

27         reverse(s, 0, len - 1);

28         // reverse every word

29         i = 0;

30         while (i < len) {

31             j = i;

32             while (j < len && s[j] != ' ') {

33                 ++j;

34             }

35             reverse(s, i, j - 1);

36             i = j;

37             while (i < len && s[i] == ' ') {

38                 ++i;

39             }

40         }

41         

42         // remove redundant spaces between words

43         offset = 0;

44         i = 0;

45         while (true) {

46             j = i;

47             while (j < len && s[j] != ' ') {

48                 s[j - offset] = s[j];

49                 ++j;

50             }

51             i = j;

52             if (i == len) {

53                 break;

54             }

55             s[i - offset] = s[i];

56             ++i;

57             while (i < len && s[i] == ' ') {

58                 ++i;

59                 ++offset;

60             }

61         }

62         

63         while (offset > 0) {

64             s.pop_back();

65             --offset;

66         }

67     }

68 private:

69     void reverse(string &s, int ll, int rr) {

70         int i;

71         char ch;

72         

73         for (i = ll; i < ll + rr - i; ++i) {

74             ch = s[i];

75             s[i] = s[ll + rr - i];

76             s[ll + rr - i] = ch;

77         }

78     }

79 };