Welcome to Subscribe On Youtube

Question

Formatted question description: https://leetcode.ca/all/186.html

Given a character array s, reverse the order of the words.

A word is defined as a sequence of non-space characters. The words in s will be separated by a single space.

Your code must solve the problem in-place, i.e. without allocating extra space.

 

Example 1:

Input: s = ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
Output: ["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]

Example 2:

Input: s = ["a"]
Output: ["a"]

 

Constraints:

  • 1 <= s.length <= 105
  • s[i] is an English letter (uppercase or lowercase), digit, or space ' '.
  • There is at least one word in s.
  • s does not contain leading or trailing spaces.
  • All the words in s are guaranteed to be separated by a single space.

Algorithm

First flip each word, then the entire string,

Or you can reverse the order, first flip the entire string, and then flip each word.

Code

  • 
public class Reverse_Words_in_a_String_II {

    public void reverseWords(char[] s) {
        reverse(s, 0, s.length);

        int l = 0; // left
        int r = 0; // right
        while (r <= s.length) {
            if (r == s.length || s[r] == ' ') {
                reverse(s, l, r);
                l = r + 1; // +1 to skip space
            }

            r++;
        }
    }

    // end is exclusive
    private void reverse(char[] s, int begin, int end) {
        for (int i = 0; i < (end - begin) / 2; i++) {
            char temp = s[begin + i];
            s[begin + i] = s[end - i - 1];
            s[end - i - 1] = temp;
        }
    }

}

############

class Solution {
    public void reverseWords(char[] s) {
        int n = s.length;
        for (int i = 0, j = 0; j < n; ++j) {
            if (s[j] == ' ') {
                reverse(s, i, j - 1);
                i = j + 1;
            } else if (j == n - 1) {
                reverse(s, i, j);
            }
        }
        reverse(s, 0, n - 1);
    }

    private void reverse(char[] s, int i, int j) {
        for (; i < j; ++i, --j) {
            char t = s[i];
            s[i] = s[j];
            s[j] = t;
        }
    }
}

  • // OJ: https://leetcode.com/problems/reverse-words-in-a-string-ii/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    void reverseWords(vector<char>& s) {
        int i = 0, N = s.size();
        while (i < N) {
            int start = i;
            while (i < N && s[i] != ' ') ++i;
            reverse(begin(s) + start, begin(s) + i);
            ++i;
        }
        reverse(begin(s), end(s));
    }
};

  • class Solution:
    def reverseWords(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """

        def reverse(s, i, j):
            while i < j:
                s[i], s[j] = s[j], s[i]
                i += 1
                j -= 1

        i, j, n = 0, 0, len(s)
        while j < n:
            if s[j] == ' ':
                reverse(s, i, j - 1)
                i = j + 1
            elif j == n - 1:
                reverse(s, i, j)
            j += 1
        reverse(s, 0, n - 1)

############

class Solution:
  # @param s, a list of 1 length strings, e.g., s = ['h','e','l','l','o']
  # @return nothing
  def reverseWords(self, s):
    def swap(start, end, slist):
      while start < end:
        slist[start], slist[end] = slist[end], slist[start]
        start += 1
        end -= 1

    wstart, wend = 0, 0
    for i in range(0, len(s)):
      if s[i] == " ":
        wend = i - 1
        swap(wstart, wend, s)
        wstart = i + 1
      elif i + 1 == len(s):
        swap(wstart, i, s)

    swap(0, len(s) - 1, s)


  • func reverseWords(s []byte) {
	n := len(s)
	for i, j := 0, 0; j < n; j++ {
		if s[j] == ' ' {
			reverse(s, i, j-1)
			i = j + 1
		} else if j == n-1 {
			reverse(s, i, j)
		}
	}
	reverse(s, 0, n-1)
}

func reverse(s []byte, i, j int) {
	for i < j {
		s[i], s[j] = s[j], s[i]
		i++
		j--
	}
}

All Problems

All Solutions