力扣每日一题2022-09-07简单题：重新排列单词间的空格

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题目描述

重新排列单词间的空格

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思路

模拟

题目给定字符串text，按照空格分割，得到单词集合，并统计空格数。

• 如果只有1个单词，则将所有空格拼接放到单词末尾即可。
• 否则计算出单词间的间隔，并按照单词及间隔来进行拼接，若拼接后还有多余的空格，则拼接到字符串末尾。

Python实现

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
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Java实现

class Solution {
    public String reorderSpaces(String text) {
        int length = text.length();
        String[] words = text.trim().split("\\s+");
        int space = length;
        for (String word : words) {
            space -= word.length();
        }
        StringBuilder sb = new StringBuilder();
        if (words.length == 1) {
            sb.append(words[0]);
            for (int i = 0; i < space; i++) {
                sb.append(' ');
            }
            return sb.toString();
        }
        int perSpace = space / (words.length - 1);
        int restSpace = space % (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();
    }
}
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