找出给定数组中第一个不重复的数组元素

1. 不用Java的API，只是利用循环查找需要的数据元素。

public class FindUnique {

	private static void findUniqueMember(char[] array){
		
                System.out.print("The array includes: ");
	        System.out.println(array);

		for (int i=0; i<array.length; i++){
		  for (int j=0; j<array.length; j++){
		    if (i != j){
			   if (array[i] == array[j]){
				System.out.println ("Member: " + i + " [" + array[i] + "] equals with " + "Member: [" + j + "] "+ array[j]);
				break;
			   }
			   else if ((array[i] != array[j])&&(j == array.length-1)){
				System.out.println ("Member: " + i + " [" + array[i] + "] is the fist unique member.\n");
				return;
			   }
		   }
		  }
		}
		System.out.println("There is no unique member in this array!\n");
	}	
	
	public static void main (String args[]){
		char[] array1 = {'a', 'b', 'c', 'd', 'e', 'f'};
		char[] array2 = {'a', 'a', 'd', 'd', 'e', 'f'};
		char[] array3 = {'a', 'b', 'c', 'd', 'e', 'a'};
		char[] array4 = {'a', 'a', 'a', 'a', 'a', 'a'};
		
		findUniqueMember (array1);
		findUniqueMember (array2);
		findUniqueMember (array3);
		findUniqueMember (array4);
	}
}

结果是：

The array includes: abcdef
Member: 0 [a] is the fist unique member.

The array includes: aaddef
Member: 0 [a] equals with Member: [1] a
Member: 1 [a] equals with Member: [0] a
Member: 2 [d] equals with Member: [3] d
Member: 3 [d] equals with Member: [2] d
Member: 4 [e] is the fist unique member.

The array includes: abcdea
Member: 0 [a] equals with Member: [5] a
Member: 1 [b] is the fist unique member.

The array includes: aaaaaa
Member: 0 [a] equals with Member: [1] a
Member: 1 [a] equals with Member: [0] a
Member: 2 [a] equals with Member: [0] a
Member: 3 [a] equals with Member: [0] a
Member: 4 [a] equals with Member: [0] a
Member: 5 [a] equals with Member: [0] a

2. 如何利用Java的API来实现呢？

public class FindUniqueAPI {
	public static void findUniqueMember (char[] array){
		String arrayString = "";
		
		for (int i=0; i<array.length; i++){
		   arrayString += array[i];
		}
		
		System.out.println("The arrayString is: "+ arrayString);
		
		for (int j=0; j<array.length; j++){
		   String tempString = arrayString.replaceAll(array[j]+"", "");
			
		   if ((arrayString.length()-tempString.length()) == 1){
		      System.out.println ("The unique member is: "+ array[j] +"\n");
		      return;
		   }
		}
		System.out.println("There is no unique member in this array!\n");
	}
	
	public static void main (String args[]){
	   char[] array1 = {'a', 'b', 'c', 'd', 'e', 'f'};
	   char[] array2 = {'a', 'a', 'd', 'd', 'e', 'f'};
	   char[] array3 = {'a', 'b', 'c', 'd', 'e', 'a'};
	   char[] array4 = {'a', 'a', 'a', 'a', 'a', 'a'};
		
	   findUniqueMember (array1);
	   findUniqueMember (array2);
	   findUniqueMember (array3);
	   findUniqueMember (array4);
	}
}

结果是：

The arrayString is: abcdef
The unique member is: a

The arrayString is: aaddef
The unique member is: e

The arrayString is: abcdea
The unique member is: b

The arrayString is: aaaaaa
There is no unique member in this array!

不知道还有没有别的更好的方法...