Reverse Words in a String–LeetCode#151

151.Reverse Words in a String

Given an input string, reverse the string word by word.
For example,
Given s = “the sky is blue“,
return “blue is sky the“.
思路:给定一个字符串，逆向排序后输出。可以转换成字符串数组排序后输出，注意的是String.split()函数，如果给定的字符串中有多个空格，如：

"   a   b "

如果单单使用String.split(” “)会返回中间的空格，所以需要使用String.split(” +”);
代码如下:

/**
 * Created by Poldi on 2017/7/7.
 */
public class ReverseWordsInAString {
    public String reverseWords(String s) {
        String[] strings = s.trim().split(" +");
        String result = "";
        for (int i=0;i<strings.length/2;i++){
            String temp = strings[i];
            strings[i] = strings[strings.length-1-i];
            strings[strings.length-1-i] = temp;
        }
        result = String.join(" ",strings);

        return result;
    }
    public static void main(String[] args) {
        ReverseWordsInAString reverseWordsInAString = new ReverseWordsInAString();
        System.out.println(reverseWordsInAString.reverseWords("   a   b "));
    }
}

这边我是直接运用了循环遍历，前后对应赋值的方法，还有更简便的方法：

public String reverseWords(String s) {
    String[] strings = s.trim().split(" +");
    Collections.reverse(Arrays.asList(strings));
    String result = String.join(" ",strings);
    return result;
}

直接运用了Collections的reverse方法

@SuppressWarnings({"rawtypes", "unchecked"})
public static void reverse(List<?> list) {
    int size = list.size();
    if (size < REVERSE_THRESHOLD || list instanceof RandomAccess) {
        for (int i=0, mid=size>>1, j=size-1; i<mid; i++, j--)
            swap(list, i, j);
    } else {
        // instead of using a raw type here, it's possible to capture
        // the wildcard but it will require a call to a supplementary
        // private method
        ListIterator fwd = list.listIterator();
        ListIterator rev = list.listIterator(size);
        for (int i=0, mid=list.size()>>1; i<mid; i++) {
            Object tmp = fwd.next();
            fwd.set(rev.previous());
            rev.set(tmp);
        }
    }
}

从reverse的源码中可以看出逆序的原理是相同的