class Solution {
public:
string reorderSpaces(string text) {
}
};
1592. 重新排列单词间的空格
给你一个字符串 text ，该字符串由若干被空格包围的单词组成。每个单词由一个或者多个小写英文字母组成，并且两个单词之间至少存在一个空格。题目测试用例保证 text 至少包含一个单词 。
请你重新排列空格，使每对相邻单词之间的空格数目都 相等 ，并尽可能 最大化 该数目。如果不能重新平均分配所有空格，请 将多余的空格放置在字符串末尾 ，这也意味着返回的字符串应当与原 text 字符串的长度相等。
返回 重新排列空格后的字符串 。
示例 1：
输入：text = " this is a sentence " 输出："this is a sentence" 解释：总共有 9 个空格和 4 个单词。可以将 9 个空格平均分配到相邻单词之间，相邻单词间空格数为：9 / (4-1) = 3 个。
示例 2：
输入：text = " practice makes perfect" 输出："practice makes perfect " 解释：总共有 7 个空格和 3 个单词。7 / (3-1) = 3 个空格加上 1 个多余的空格。多余的空格需要放在字符串的末尾。
示例 3：
输入：text = "hello world" 输出："hello world"
示例 4：
输入：text = " walks udp package into bar a" 输出："walks udp package into bar a "
示例 5：
输入：text = "a" 输出："a"
提示：
1 <= text.length <= 100text 由小写英文字母和 ' ' 组成text 中至少包含一个单词原站题解
golang 解法, 执行用时: 0 ms, 内存消耗: 1.8 MB, 提交时间: 2023-09-27 15:20:03
func reorderSpaces(s string) (ans string) {
words := strings.Fields(s)
space := strings.Count(s, " ")
lw := len(words) - 1
if lw == 0 {
return words[0] + strings.Repeat(" ", space)
}
return strings.Join(words, strings.Repeat(" ", space/lw)) + strings.Repeat(" ", space%lw)
}
javascript 解法, 执行用时: 60 ms, 内存消耗: 40.8 MB, 提交时间: 2023-09-27 15:19:41
/**
* @param {string} text
* @return {string}
*/
var reorderSpaces = function(text) {
const length = text.length;
const words = [];
text.split(' ').forEach(e => {
if (e.length > 0) {
words.push(e);
}
});
let cntSpace = length;
for (const word of words) {
if (word.length) {
cntSpace -= word.length;
}
}
let sb = '';
if (words.length === 1) {
sb += words[0];
for (let i = 0; i < cntSpace; i++) {
sb += ' ';
}
return sb;
}
const perSpace = Math.floor(cntSpace / (words.length - 1));
const restSpace = cntSpace % (words.length - 1);
for (let i = 0; i < words.length; i++) {
if (i > 0) {
for (let j = 0; j < perSpace; j++) {
sb += ' ';
}
}
sb += words[i];
}
for (let i = 0; i < restSpace; i++) {
sb += ' ';
}
return sb;
};
cpp 解法, 执行用时: 4 ms, 内存消耗: 6.6 MB, 提交时间: 2023-09-27 15:19:25
class Solution {
public:
vector<string_view> split(const string_view &str, char trim) {
int n = str.size();
vector<string_view> res;
int pos = 0;
while (pos < n) {
while(pos < n && str[pos] == trim) {
pos++;
}
if (pos < n) {
int curr = pos;
while(pos < n && str[pos] != trim) {
pos++;
}
res.emplace_back(str.substr(curr, pos - curr));
}
}
return res;
}
string reorderSpaces(string text) {
int length = text.size();
vector<string_view> words = split(text, ' ');
int cntSpace = length;
int wordCount = 0;
for (auto &word : words) {
if (word.size() > 0) {
cntSpace -= word.size();
wordCount++;
}
}
string ans;
if (words.size() == 1) {
ans.append(words[0]);
for (int i = 0; i < cntSpace; i++) {
ans.push_back(' ');
}
return ans;
}
int perSpace = cntSpace / (wordCount - 1);
int restSpace = cntSpace % (wordCount - 1);
for (int i = 0; i < words.size(); i++) {
if (words[i].size() == 0) {
continue;
}
if (ans.size() > 0) {
for (int j = 0; j < perSpace; j++) {
ans.push_back(' ');
}
}
ans.append(words[i]);
}
for (int i = 0; i < restSpace; i++) {
ans.push_back(' ');
}
return ans;
}
};
java 解法, 执行用时: 3 ms, 内存消耗: 39.8 MB, 提交时间: 2023-09-27 15:19:11
class Solution {
public String reorderSpaces(String text) {
int length = text.length();
String[] words = text.trim().split("\\s+");
int cntSpace = length;
for (String word : words) {
cntSpace -= word.length();
}
StringBuilder sb = new StringBuilder();
if (words.length == 1) {
sb.append(words[0]);
for (int i = 0; i < cntSpace; i++) {
sb.append(' ');
}
return sb.toString();
}
int perSpace = cntSpace / (words.length - 1);
int restSpace = cntSpace % (words.length - 1);
for (int i = 0; i < words.length; i++) {
if (i > 0) {
for (int j = 0; j < perSpace; j++) {
sb.append(' ');
}
}
sb.append(words[i]);
}
for (int i = 0; i < restSpace; i++) {
sb.append(' ');
}
return sb.toString();
}
}
python3 解法, 执行用时: 28 ms, 内存消耗: 16 MB, 提交时间: 2023-09-27 15:18:52
class Solution:
def reorderSpaces(self, text: str) -> str:
words = text.split()
space = text.count(' ')
if len(words) == 1:
return words[0] + ' ' * space
per_space, rest_space = divmod(space, len(words) - 1)
return (' ' * per_space).join(words) + ' ' * rest_space
golang 解法, 执行用时: 0 ms, 内存消耗: 2.2 MB, 提交时间: 2021-06-16 17:12:52
func reorderSpaces(text string) string {
blank := 0
for i := range text {
if text[i] == ' ' {
blank++
}
}
re, _ := regexp.Compile("\\s+")
text = strings.TrimSpace(text)
text = re.ReplaceAllString(text, " ")
words := strings.Split(text, " ")
if len(words) == 1 {
return text + strings.Repeat(" ", blank)
}
r := blank % (len(words) - 1)
ans := strings.Join(words, strings.Repeat(" ", blank/(len(words)-1)))
if r != 0 {
ans += strings.Repeat(" ", r)
}
return ans
}