9.7 Reverse Words in a String

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2023-12-01

Question

leetcode: Reverse Words in a String
lintcode: Reverse Words in a String

Problem Statement

Given an input string, reverse the string word by word.

For example,
Given s = "the sky is blue",
return "blue is sky the".

Update (2015-02-12):
For C programmers: Try to solve it in-place in O(1) space.

Clarification:

What constitutes a word?
A sequence of non-space characters constitutes a word.
Could the input string contain leading or trailing spaces?
Yes. However, your reversed string should not contain leading or trailing spaces.
How about multiple spaces between two words?
Reduce them to a single space in the reversed string.

题解

由第一个提问可知：题中只有空格字符和非空格字符之分，因此空格字符应为其一关键突破口。
由第二个提问可知：输入的前导空格或者尾随空格在反转后应去掉。
由第三个提问可知：两个单词间的多个空格字符应合并为一个或删除掉。

首先找到各个单词(以空格隔开)，根据题目要求，单词应从后往前依次放入。split 后从后往前加入空格返回即可。如果不使用 split 的话正向取出比较麻烦，因此可尝试采用逆向思维——先将输入字符串数组中的单词从后往前逆序取出，取出单词后即翻转并append至新字符串数组。在append之前加入空格即可，即两次翻转法。

Python

class Solution(object):
    def reverseWords(self, s):
        """
        :type s: str
        :rtype: str
        """
        return ' '.join(reversed(s.strip().split()))

C++

class Solution {
public:
    /**
     * @param s : A string
     * @return : A string
     */
    string reverseWords(string s) {
        if (s.empty()) {
            return s;
        }

        string s_ret, s_temp;
        string::size_type ix = s.size();
        while (ix != 0) {
            s_temp.clear();
            while (!isspace(s[--ix])) {
                s_temp.push_back(s[ix]);
                if (ix == 0) {
                    break;
                }
            }
            if (!s_temp.empty()) {
                if (!s_ret.empty()) {
                    s_ret.push_back(' ');
                }
                std::reverse(s_temp.begin(), s_temp.end());
                s_ret.append(s_temp);
            }
        }

        return s_ret;
    }
};

Java

public class Solution {
    public String reverseWords(String s) {
        if (s == null || s.trim().isEmpty()) {
            return "";
        }

        String[] words = s.split(" ");
        StringBuilder sb = new StringBuilder();
        for (int i = words.length - 1; i >= 0; i--) {
            if (!words[i].isEmpty()) {
                sb.append(words[i]).append(" ");
            }
        }

        return sb.substring(0, sb.length() - 1);
    }
}

源码分析

首先处理异常，s 为空或者空白字符串时直接返回空。
如果首先排除掉空白字符串则后面不需要为长度为0单独考虑。
Java 中使用 StringBuilder 效率更高。

空间复杂度为 O(1) 的解法？

处理异常及特殊情况
处理多个空格及首尾空格
记住单词的头尾指针，翻转之
整体翻转