题目：输入一个整数数组，实现一个函数来调整该数组中数字的属性怒，使得所有奇数位于数组的前半部分，所有偶数位于数组的后半部分。
思路：其实就是用快速排序法的第一轮排序，从左右夹逼，左边遇到偶数，停下来，右边遇到奇数，停下来，交换，再夹逼，直到两个指针相遇。
代码实现：
package com.yyq; /**
* Created by Administrator on 2015/9/13.
*/
public class ReorderArray {
public static void reorderOddEven(int[] data){
if(data == null) return;
int start = 0;
int end = data.length - 1;
//奇数放前面，偶数放后面
while(start < end){
while(start < end && (data[start] & 0x1) != 0){
start++;
}
while(start < end && (data[end] & 0x1) == 0){
end--;
}
int temp = data[start];
data[start] = data[end];
data[end] = temp;
}
} // ====================测试代码====================
public static void printArray(int numbers[])
{
if(numbers == null)
return;
int len = numbers.length;
for(int i = 0; i < len; ++i)
System.out.print(numbers[i]+"\t");
System.out.println();
} public static void Test(String testName, int numbers[])
{
if(testName != null)
System.out.println(testName+" begins:");
if (numbers == null)
return;
System.out.println("Test for solution 1:");
printArray(numbers);
reorderOddEven(numbers);
printArray(numbers);
System.out.println();
} public static void Test1()
{
int numbers[] = {1, 2, 3, 4, 5, 6, 7};
Test("Test1", numbers);
} public static void Test2()
{
int numbers[] = {2, 4, 6, 1, 3, 5, 7};
Test("Test2", numbers);
} public static void Test3()
{
int numbers[] = {1, 3, 5, 7, 2, 4, 6};
Test("Test3", numbers);
} public static void Test4()
{
int numbers[] = {1};
Test("Test4", numbers);
} public static void Test5()
{
int numbers[] = {2};
Test("Test5", numbers);
} public static void Test6()
{
Test("Test6", null);
} public static void main(String[] args)
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
}
}
输出结果：
Test1 begins:
Test for solution 1:
1 2 3 4 5 6 7
1 7 3 5 4 6 2
Test2 begins:
Test for solution 1:
2 4 6 1 3 5 7
7 5 3 1 6 4 2
Test3 begins:
Test for solution 1:
1 3 5 7 2 4 6
1 3 5 7 2 4 6
Test4 begins:
Test for solution 1:
1
1
Test5 begins:
Test for solution 1:
2
2
Test6 begins: