求一个数组的算法

[雨松MOMO]

有2个String数组 长度一摸一样 求一个算法 先在name中把相同的字符串合并成一个，然后在num中也相应的合并。

例如

//举一个例子 比如现在有2个数组
String []name ={"aa","aa","bb","bb","cc"} //这只是一个范例 因为程序中调用的时候数组元素是会变化的
String []num ={"123","234","235","543","145"} //这只是一个范例 因为程序中调用的时候数组元素是会变化的

//我想要的结果是
String []name = {"aa","bb","cc"};
String []num = {"123,234", "235,543", "145"};

[qybao]

原来LZ是J2ME，那是没有map和list的，可以用HashTable

import java.util.*;

public class Test {
    public static void main(String[] args) {
        try {
            String []name ={"aa","aa","bb","bb","cc"};
            String []num ={"123","234","235","543","145"};
            HashTable map = new HashTable();
            for (int i=0; i<name.length; i++) {
                Object v = map.get(name[i]);
                if (v == null) {
                    map.put(name[i], num[i]);
                } else {
                    map.put(name[i], v+","+num[i]);
                }
            }
            name = new String[map.size()];
            for (int k=0, Enumeration e = map.keys(); e.hasMoreElements(); k++) {
                name[k] = String.valueOf(e.nextElement());
                System.out.print(name[k]+"\t");
            }
            num = new String[map.size()];
            for (int k=0, Enumeration e = map.elements(); e.hasMoreElements(); k++) {
                num[k] = String.valueOf(e.nextElement());
                System.out.print(num[k]+"\t");
            }
        } catch (Throwable e) {
            e.printStackTrace();
        }
    }
}

[Inhibitory]

import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

public class Test {
    public static void main(String[] args) {
        String []name ={"aa","aa","bb","bb","cc"}; //这只是一个范例 因为程序中调用的时候数组元素是会变化的
        String []num ={"123","234","235","543","145"}; //这只是一个范例 因为程序中调用的时候数组元素是会变化的
        
        Set<String> set = new HashSet<String>();
        for (String n : name) {
            set.add(n);
        }
        
        name = set.toArray(new String[0]);
        System.out.println(Arrays.toString(name));
    }
}

[Inhibitory]

import java.util.Arrays;
import java.util.HashSet;

public class Test {
    public static void main(String[] args) {
        String []name = {"aa","aa","bb","bb","cc"}; //这只是一个范例 因为程序中调用的时候数组元素是会变化的
        String []num = {"123","234","235","543","145"}; //这只是一个范例 因为程序中调用的时候数组元素是会变化的
        
        name = Test.uniqueArray(name);
        num = Test.uniqueArray(num);
        
        System.out.println(Arrays.toString(name));
        System.out.println(Arrays.toString(num));
    }
    
    // 去掉数组中重复的元素
    public static String[] uniqueArray(String[] arr) {
        return new HashSet<String>(Arrays.asList(arr)).toArray(new String[0]);
    }
}

[hmc1985]

在j2me里面，用Hashtable和Vector都可以实现这个功能

[mouhk]

用一个哈希表来记录数组name是否重复，遍历处理其中的元素就是了，如果重复，则对应的处理num。我觉得这个算法思路还是很明确的三。

[雨松MOMO]

高手速度来

[qybao]

String []name ={"aa","aa","bb","bb","cc"};
String []num ={"123","234","235","543","145"};
List<String> nameList = new ArrayList<String>();
List<String> numList = new ArrayList<String>();
for (String s : name) {
    if (nameList.contains(s)) {
        continue;
    }
    nameList.add(s);
    int idx = Arrays.binarySearch(name, 0, name.length, s);
    StringBuilder sb = new StringBuilder(num[idx]);
    while (true) {
        idx = Arrays.binarySearch(name, idx+1, name.length, s);
        if (idx < 0) {
            break;
        }
        sb.append(",").append(num[idx]);
    }
    numList.add(sb.toString());
}
name = nameList.toArray(new String[0]);
num = numList.toArray(new String[0]);
System.out.println(Arrays.toString(name));
System.out.println(Arrays.toString(num));

[雨松MOMO]

楼上的编译不通过

[雨松MOMO]

高手快来啊 我在线等啊

[e9876]

2楼的应该可以了吧。

[雨松MOMO]

高手快来啊！！！

[雨松MOMO]

高手快来啊

[雨松MOMO]

高手在那里啊····

[qybao]

[Quote=引用 8 楼 xys289187120 的回复:]
楼上的编译不通过
[/Quote]
怎么会编译通不过了？
jdk多少的？程序都写到这份上，编译出问题就不会自己找原因吗？
以下测试通过，跟和上面代码完全一样，输出结果
[aa, bb, cc]
[123,234, 235,543, 145]

import java.util.*;

public class Test {
    public static void main(String[] args) {
        try {
            String []name ={"aa","aa","bb","bb","cc"};
            String []num ={"123","234","235","543","145"};
            List<String> nameList = new ArrayList<String>();
            List<String> numList = new ArrayList<String>();
            for (String s : name) {
                if (nameList.contains(s)) {
                    continue;
                }
                nameList.add(s);
                int idx = Arrays.binarySearch(name, 0, name.length, s);
                StringBuilder sb = new StringBuilder(num[idx]);
                while (true) {
                    idx = Arrays.binarySearch(name, idx+1, name.length, s);
                    if (idx < 0) {
                        break;
                    }
                    sb.append(",").append(num[idx]);
                }
                numList.add(sb.toString());
            }
            name = nameList.toArray(new String[0]);
            num = numList.toArray(new String[0]);
            System.out.println(Arrays.toString(name));
            System.out.println(Arrays.toString(num));     
        } catch (Throwable e) {
            e.printStackTrace();
        }
    }        
}

[qybao]

再给你一个用map实现的方法，这个跟上面的区别是，用map不一定能保证跟原来的数组的顺序一致
自己慢慢体会吧

import java.util.*;

public class Test {
    public static void main(String[] args) {
        try {
            String []name ={"aa","aa","bb","bb","cc"};
            String []num ={"123","234","235","543","145"};
            Map<String, String> map = new HashMap<String, String>();
            for (int i=0; i<name.length; i++) {
                String v = map.get(name[i]);
                if (v == null) {
                    map.put(name[i], num[i]);
                } else {
                    map.put(name[i], v+","+num[i]);
                }
            }
            name = map.keySet().toArray(new String[0]);
            num = map.values().toArray(new String[0]);
            System.out.println(Arrays.toString(name));
            System.out.println(Arrays.toString(num));
        } catch (Throwable e) {
            e.printStackTrace();
        }
    }
}

[雨松MOMO]

高手速度来。。