java实现字母排序，区分大小写

dashan70345 2010-09-09 03:10:52

实现一个算法对多个字符串进行排序，不能使用现有的类，在排序中，字符串“Bc”，“Ad”，“aC”，“Hello”，“Xman”，“little”，“During”，“day”能够排序成“Ad”，“aC”，“Bc”，“During”，“day”，“Hello”，“little”，“Xman”，也就是说，在排序的过程中并不是传统的按照字符串排序，在排序中海需要将小写字母一并排序，也就是说a字符串要在B和b之前

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zhuzeitou 2010-09-13

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len是s1、s2中较短者的长度，如果能走到这一步，说明两个字符串的前len个字符完全相同
所以这一段代码是对s1是s2的子串或s2是s1的子串的情况下做处理
在这个前提下，如果len1==len2，说明两个字符串的确相同；如果len1<len2，那么s1是s2的子串，s1<s2；如果len1>len2，那么s2是s1的子串，s1>s2
其实这里我定了默认条件都是英文字符，如果有其他字符可能会有问题

dashan70345 2010-09-11

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12楼的有问题，之比较了首字母
String[] s = {"Bc","Ad","aa","AC","Hello","Xman","little","Da","During","dfy"};
就会出错

dashan70345 2010-09-11

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15楼
compare(String s1, String s2)中的这段代码不懂，帮忙解释下
if (len1 == len2)
return 0;
else if (len1 > len2)
return arr1[len];
else
return -arr2[len];
谢了。

zhuzeitou 2010-09-10

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public static void sort(String[] src) {

    int[][] arr = new int[src.length][2];

    int count = 0, index = 0;

    arr[0][0] = 0;

    arr[0][1] = src.length - 1;

    while (index <= count) {

        int left = arr[index][0];

        int right = arr[index][1];

        int[] dir = { 0, 1 };

        int i = left, j = right;

        String temp;

        while (i < j) {

            if (compare(src[i], src[j]) > 0) {

                temp = src[i];

                src[i] = src[j];

                src[j] = temp;

                dir[0] = 1 - dir[0];

                dir[1] = 1 - dir[1];

            }

            i += dir[0];

            j -= dir[1];

        }

        if (j - left > 1) {

            count++;

            arr[count][0] = left;

            arr[count][1] = j - 1;

        }

        if (right - i > 1) {

            count++;

            arr[count][0] = i + 1;

            arr[count][1] = right;

        }

        index++;

    }

}

改一个排序算法……#15的是冒泡，这个是无递归快排……

zhuzeitou 2010-09-10

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public static int compare(String s1, String s2) {

    char[] arr1 = s1.toCharArray(), arr2 = s2.toCharArray();

    int index = 0, len1 = arr1.length, len2 = arr2.length;

    int len = len1 < len2 ? len1 : len2;

    while (index < len) {

        char c1 = arr1[index], c2 = arr2[index];

        char c1_ = (char) (c1 >= 'a' ? c1 - ('a' - 'A') : c1);

        char c2_ = (char) (c2 >= 'a' ? c2 - ('a' - 'A') : c2);

        if (c1_ == c2_) {

            if (c1 != c2)

                return c1 - c2;

        } else

            return c1_ - c2_;

        index++;

    }

    if (len1 == len2)

        return 0;

    else if (len1 > len2)

        return arr1[len];

    else

        return -arr2[len];

}



public static void sort(String[] src) {

    String temp;

    for (int i = 0; i < src.length - 1; i++) {

        for (int j = 0; j < src.length - i - 1; j++) {

            if (compare(src[j], src[j + 1]) > 0) {

                temp = src[j];

                src[j] = src[j + 1];

                src[j + 1] = temp;

            }

        }

    }

}



public static void main(String[] args) {

    String[] s = { "Bc", "Ad", "aC", "Hello", "Xman", "little", "During",

            "day" };

    System.out.println("Before sort: " + Arrays.toString(s));

    sort(s);

    System.out.println("After sort: " + Arrays.toString(s));

}

bf234511171 2010-09-10

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11楼你的答案有问题吧LZ说了不能使用已有的类你写的那个类明显用倒了已有的类了！

ytyjxhy 2010-09-10

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public class Test {
public static void main(String[] args) {
String[] s = {"Bc","Ad","aC","Hello","Xman","little","During","day"};
String s1;
for(int i =0;i<s.length;i++){
for(int j=i;j<s.length;j++){
if(s[i].compareToIgnoreCase(s[j])>0){
s1=s[i];
s[i]=s[j];
s[j]=s1;
}
}
}
String ss;
for(ss:s){
System.out.println(ss);
}
}
}

tong467 2010-09-09

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public class Test2 {
public static void main(String[] args) {
String[] s = {"Bc","Ad","aC","Hello","Xman","little","During","day"};
for(int i =0;i<s.length;i++){
for(int j=i;j<s.length;j++){
if(px(s[i].charAt(0),s[j].charAt(0))){
String aa = "";
aa = s[i];
s[i] = s[j];
s[j] = aa;
}
}
}
for(String k: s){
System.out.println(k);
}
}
private static boolean px(char c1,char c2){
boolean b = true;
if(c1>=97){
c1 -= 32;
}if(c2>=97){
c2 -= 32;
}
if(c1==c2&&c1<=c2){
b = false;
}
if(c1<c2){
b = false;
}
return b;
}
}
刚刚那个逻辑有问题这个好像比那个强点！

tong467 2010-09-09

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package com;

public class Test2 {
public static void main(String[] args) {
String[] s = {"Bc","Ad","aC","Hello","Xman","little","During","day"};
for(int i =0;i<s.length;i++){
for(int j=i;j<s.length;j++){
if(px(s[i].charAt(0),s[j].charAt(0))){
String aa = "";
aa = s[i];
s[i] = s[j];
s[j] = aa;
}
}
}
for(String k: s){
System.out.println(k);
}
}
private static boolean px(char c1,char c2){
boolean b = true;
if(c1>=97){
c1 -= 32;
}if(c2>=97){
c2 -= 32;
}
if(c1<=c2){
b = false;
}
return b;
}
}
你看下吧我英语次变量名和方法名很难认. 呵呵！包含下！

tong467 2010-09-09