字符串翻转

1.编写程序，在原字符串中把字符串尾部的m个字符移动到字符串的头部，要求：长度为n的字符串操作时间复杂度为O(n)，空间复杂度为O(1)。 例如，原字符串为”Ilovebaofeng”，m=7，输出结果为：”baofengIlove”。

2、单词翻转。输入一个英文句子，翻转句子中单词的顺序，但单词内字符的顺序不变，句子中单词以空格符隔开。为简单起见，标点符号和普通字母一样处理。例如，输入“I am a student.”，则输出“student. a am I”。

代码实现：

package alg;

/**
 * @author zha
 *  字符串反转
 */
public class Alg1StringReserv {
    
    public static void main(String[] args) {
        String test = "ilovebaofeng";
        int index = 7 ;
        String re = reserve(test,index);
        System.out.println(re);
        
        String test2 = "I am a student.";
        char index2 = ' ';
        String re2 = reserve(test2,index2);
        System.out.println(re2);

    }
    
    //根据某一个特定的字符进行翻转
    private static String reserve(String test, char index) {
        
        char[] chars = test.toCharArray();
        int from = 0,to = 0;
        for (int i = 0; i < chars.length; i++) {
            char temp = chars[i];
            if(temp == index){
                to = i-1;
                turnOver(chars,from,to);
                from = i+1;
            }
        }
        
        if(to < chars.length - 1){
            turnOver(chars, from, chars.length-1);
        }
        
        turnOver(chars,0,chars.length-1);
        return new String(chars);
    }

    private static String reserve(String test, int index) {
        int length = test.length();
        int begine = （length - index - 1）%length;
        char[] chars = test.toCharArray();
        turnOver(chars,0,begine);
        turnOver(chars,begine+1,length-1);
        turnOver(chars,0,length-1);
        return new String(chars);
    }

    private static void turnOver(char[] chars, int begine, int to) {
        if(begine < to){
            for (; begine < to; begine++,to--) {
                swap(chars,begine,to);
            }
        }
        
    }

    private static void swap(char[] chars, int begine, int to) {
        char temp = chars[begine];
        chars[begine] = chars[to];
        chars[to] = temp;
        
    }

}

欢҉迎҉指҉出҉代҉码҉中҉不҉足҉的҉地҉方҉。