# 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--
}
}