有道算法题是“单词反转”，闲着没事就撸一把，动动脑~
此题有两种理解：1. 将输入字符串按字符反转输出。2. 将句子以单词为最小单位（空格区分）反转输出。
举个栗子，直观点。

输入字符串： i am a coder.
理解1，输出：.redoc a ma i
理解2，输出：.coder a am i

直接贴代码吧

// 题目： 单词反转
    public static void main(String[] args) {
        // 字符反转思路：1. 反向遍历输出 2. 利用对称性遍历一半。
        // 反转句子但不反转其中单词思路：首先剔除前后多余空格，然后反转整个句子，最后遍历反转每个单词。

        String str = "abcdef";
        String sen = " i am a student dfdkfj";

        ReverseWord reverseWord = new ReverseWord();
        String ret = reverseWord.reverseTwo(str);
        System.out.println(ret);
        System.out.println(reverseWord.reverseSentence(sen));
    }

    /**
     * 字符反转方式1
     *
     * @param originStr
     * @return
     */
    private String reverseOne(String originStr) {
        if (originStr == null || originStr.isEmpty()) {
            return originStr;
        }
        final int length = originStr.length();

        StringBuilder sBuilder = new StringBuilder(length);
        for (int i = length - 1; i >= 0; i--) {
            sBuilder.append(originStr.charAt(i));
        }
        return sBuilder.toString();
    }

    /**
     * 字符反转方式1
     *
     * @param originStr
     * @return
     */
    private String reverseTwo(String originStr) {
        if (originStr == null || originStr.isEmpty()) {
            return originStr;
        }
        final int length = originStr.length();
        char[] chars = originStr.toCharArray();
        final int halfLen = length / 2;
        for (int i = 0; i < halfLen; i++) {
            char temp = chars[i];
            chars[i] = chars[length - 1 - i];
            chars[length - 1 - i] = temp;
        }
        return new String(chars);
    }

    /**
     * 反转句子
     * @param sentence
     * @return
     */
    private String reverseSentence(String sentence) {
        if (sentence == null || sentence.isEmpty()) {
            return sentence;
        }

        sentence = sentence.trim();
        String[] strings = sentence.split(" ");
        int halfLen = strings.length / 2;
        for (int i = 0; i < halfLen; i++) {
            String temp = strings[i];
            strings[i] = strings[strings.length - 1 - i];
            strings[strings.length - 1 - i] = temp;
        }

        StringBuilder stringBuilder = new StringBuilder();
        for (int i = 0; i < strings.length; i++) {
            stringBuilder.append(strings[i]);
            stringBuilder.append(" ");
        }
        return stringBuilder.toString();
    }

写完了，思路在注释里。但总觉得实现有待优化，特别是反转句子的实现，感觉使用了sentence.split(" ")，正则匹配较耗时，有好的解题思路还请指教，欢迎浏览交流~