Description#
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.
Solutions#
Solution 1#
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| 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)
|
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| 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;
}
}
}
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| class Solution {
public:
void reverseWords(vector<char>& s) {
int n = s.size();
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);
}
void reverse(vector<char>& s, int i, int j) {
for (; i < j; ++i, --j) {
swap(s[i], s[j]);
}
}
};
|
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| 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--
}
}
|