| comments | difficulty | edit_url |
|---|---|---|
true |
简单 |
输入一个英文句子，翻转句子中单词的顺序，但单词内字符的顺序不变。为简单起见，标点符号和普通字母一样处理。例如输入字符串"I am a student. "，则输出"student. a am I"。
示例 1：
输入: "the sky is blue" 输出: "blue is sky the"
示例 2：
输入: " hello world! " 输出: "world! hello" 解释: 输入字符串可以在前面或者后面包含多余的空格，但是反转后的字符不能包括。
示例 3：
输入: "a good example" 输出: "example good a" 解释: 如果两个单词间有多余的空格，将反转后单词间的空格减少到只含一个。
说明：
- 无空格字符构成一个单词。
- 输入字符串可以在前面或者后面包含多余的空格，但是反转后的字符不能包括。
- 如果两个单词间有多余的空格，将反转后单词间的空格减少到只含一个。
注意：本题与主站 151 题相同：https://leetcode.cn/problems/reverse-words-in-a-string/
注意：此题对比原题有改动
我们可以使用双指针
时间复杂度
class Solution:
def reverseWords(self, s: str) -> str:
words = []
i, n = 0, len(s)
while i < n:
while i < n and s[i] == " ":
i += 1
if i < n:
j = i
while j < n and s[j] != " ":
j += 1
words.append(s[i:j])
i = j
return " ".join(words[::-1])class Solution {
public String reverseWords(String s) {
List<String> words = new ArrayList<>();
int n = s.length();
for (int i = 0; i < n;) {
while (i < n && s.charAt(i) == ' ') {
++i;
}
if (i < n) {
StringBuilder t = new StringBuilder();
int j = i;
while (j < n && s.charAt(j) != ' ') {
t.append(s.charAt(j++));
}
words.add(t.toString());
i = j;
}
}
Collections.reverse(words);
return String.join(" ", words);
}
}class Solution {
public:
string reverseWords(string s) {
int i = 0;
int j = 0;
int n = s.size();
while (i < n) {
while (i < n && s[i] == ' ') {
++i;
}
if (i < n) {
if (j != 0) {
s[j++] = ' ';
}
int k = i;
while (k < n && s[k] != ' ') {
s[j++] = s[k++];
}
reverse(s.begin() + j - (k - i), s.begin() + j);
i = k;
}
}
s.erase(s.begin() + j, s.end());
reverse(s.begin(), s.end());
return s;
}
};func reverseWords(s string) string {
words := []string{}
i, n := 0, len(s)
for i < n {
for i < n && s[i] == ' ' {
i++
}
if i < n {
j := i
t := []byte{}
for j < n && s[j] != ' ' {
t = append(t, s[j])
j++
}
words = append(words, string(t))
i = j
}
}
for i, j := 0, len(words)-1; i < j; i, j = i+1, j-1 {
words[i], words[j] = words[j], words[i]
}
return strings.Join(words, " ")
}function reverseWords(s: string): string {
const words: string[] = [];
const n = s.length;
let i = 0;
while (i < n) {
while (i < n && s[i] === ' ') {
i++;
}
if (i < n) {
let j = i;
while (j < n && s[j] !== ' ') {
j++;
}
words.push(s.slice(i, j));
i = j;
}
}
return words.reverse().join(' ');
}impl Solution {
pub fn reverse_words(s: String) -> String {
let mut words = Vec::new();
let s: Vec<char> = s.chars().collect();
let mut i = 0;
let n = s.len();
while i < n {
while i < n && s[i] == ' ' {
i += 1;
}
if i < n {
let mut j = i;
while j < n && s[j] != ' ' {
j += 1;
}
words.push(s[i..j].iter().collect::<String>());
i = j;
}
}
words.reverse();
words.join(" ")
}
}public class Solution {
public string ReverseWords(string s) {
List<string> words = new List<string>();
int n = s.Length;
for (int i = 0; i < n;) {
while (i < n && s[i] == ' ') {
++i;
}
if (i < n) {
System.Text.StringBuilder t = new System.Text.StringBuilder();
int j = i;
while (j < n && s[j] != ' ') {
t.Append(s[j++]);
}
words.Add(t.ToString());
i = j;
}
}
words.Reverse();
return string.Join(" ", words);
}
}class Solution {
func reverseWords(_ s: String) -> String {
var words = [String]()
var i = s.startIndex
while i < s.endIndex {
while i < s.endIndex && s[i] == " " {
i = s.index(after: i)
}
if i < s.endIndex {
var t = ""
var j = i
while j < s.endIndex && s[j] != " " {
t.append(s[j])
j = s.index(after: j)
}
words.append(t)
i = j
}
}
words.reverse()
return words.joined(separator: " ")
}
}我们可以使用语言内置的字符串分割函数，将字符串按空格分割成单词列表，然后将列表反转，再拼接成字符串即可。
时间复杂度
class Solution:
def reverseWords(self, s: str) -> str:
return " ".join(reversed(s.split()))class Solution {
public String reverseWords(String s) {
List<String> words = Arrays.asList(s.trim().split("\\s+"));
Collections.reverse(words);
return String.join(" ", words);
}
}func reverseWords(s string) string {
words := strings.Fields(s)
for i, j := 0, len(words)-1; i < j; i, j = i+1, j-1 {
words[i], words[j] = words[j], words[i]
}
return strings.Join(words, " ")
}function reverseWords(s: string): string {
return s.trim().split(/\s+/).reverse().join(' ');
}impl Solution {
pub fn reverse_words(s: String) -> String {
s.split_whitespace().rev().collect::<Vec<&str>>().join(" ")
}
}