LeetCode

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:

// 1TLE, 1AC, using another char[] is unnecessary. Don't miss '++i' or '++j'.
class Solution {
public:
    void reverseWords(string &s) {
        int i, j;
        int len;
        int offset;
        
        // remove trailing spaces
        while (s.length() > 0 && s[s.length() - 1] == ' ') {
            s.pop_back();
        }
        len = (int)s.length();
        if (len == 0) {
            return;
        }
        
        // remove leading spaces
        i = 0;
        while (i < len && s[i] == ' ') {
            ++i;
        }
        s = s.substr(i, len - i);
        len = (int)s.length();
        
        // reverse the whole string
        reverse(s, 0, len - 1);
        // reverse every word
        i = 0;
        while (i < len) {
            j = i;
            while (j < len && s[j] != ' ') {
                ++j;
            }
            reverse(s, i, j - 1);
            i = j;
            while (i < len && s[i] == ' ') {
                ++i;
            }
        }
        
        // remove redundant spaces between words
        offset = 0;
        i = 0;
        while (true) {
            j = i;
            while (j < len && s[j] != ' ') {
                s[j - offset] = s[j];
                ++j;
            }
            i = j;
            if (i == len) {
                break;
            }
            s[i - offset] = s[i];
            ++i;
            while (i < len && s[i] == ' ') {
                ++i;
                ++offset;
            }
        }
        
        while (offset > 0) {
            s.pop_back();
            --offset;
        }
    }
private:
    void reverse(string &s, int ll, int rr) {
        int i;
        char ch;
        
        for (i = ll; i < ll + rr - i; ++i) {
            ch = s[i];
            s[i] = s[ll + rr - i];
            s[ll + rr - i] = ch;
        }
    }
};