日期:2014-05-19  浏览次数:20784 次

各位大侠帮我解决这道超级简单的问题(冒泡排序的)
从控制端舒服一个字符串 比如 4848adecb 然后排序后输出结果为4488abcde.

------解决方案--------------------
public static void main(String[] args) {
String values = "4848adecb";
char[] chars = new char[values.length()];
for (int i = 0; i < chars.length; i++) {
chars[i] = values.charAt(i);
}
Arrays.sort(chars);
String value = new String(chars);
System.out.println(value);
}
------解决方案--------------------
探讨

public static void main(String[] args) {
String values = "4848adecb";
char[] chars = new char[values.length()];
for (int i = 0; i < chars.length; i++) {
chars[i] = values.charAt(i);
}
Arrays.so……

------解决方案--------------------
Java code
public static String bubbleSort(String str){
        char[] chars = new char[str.length()];
        for(int i = 0 ; i < str.length() ; i++){
            chars[i] = str.charAt(i);
        }
        for(int i = 0 ; i < chars.length ; i++){
            for(int j = i + 1 ; j < chars.length ; j ++){
                char temp;
                if(chars[i] > chars[j]){
                    temp = chars[i];
                    chars[i] = chars[j];
                    chars[j] = temp;
                }
            }
        }
        String retStr = "";
        for(int i = 0 ; i < chars.length ; i ++){
            retStr += chars[i];
        }
        return retStr;
    }

------解决方案--------------------
LZ这个是完整版的
Java code
public class BubbleSortTest {

    public static String bubbleSort(String str){
        char[] chars = new char[str.length()];
        for(int i = 0 ; i < str.length() ; i++){
            chars[i] = str.charAt(i);
        }
        for(int i = 0 ; i < chars.length ; i++){
            for(int j = i + 1 ; j < chars.length ; j ++){
                char temp;
                if(chars[i] > chars[j]){
                    temp = chars[i];
                    chars[i] = chars[j];
                    chars[j] = temp;
                }
            }
        }
        String retStr = "";
        for(int i = 0 ; i < chars.length ; i ++){
            retStr += chars[i];
        }
        return retStr;
    }
    
    public static void main(String[] args) {
        System.out.println(bubbleSort(args.length > 0 ? args[0] : "kief351"));//如果控制端没有输入字符串就默认传一个
    }
}

------解决方案--------------------
探讨

LZ这个是完整版的
Java code
public class BubbleSortTest {

public static String bubbleSort(String str){
char[] chars = new char[str.length()];
for(int i = 0 ; i < str.length() ; i++){
……