Adjust the array to make the odd number element an odd number, and the even number is an even number.

This article is entitled Baidu Quality Department interview questions

2n number, half odd number, half even number, design a program to make the number on the odd digit is odd, the number on the even bit is even, and calculate the space complexity and time complexity of the program

 

Program ideology:

This article is similar to adjusting the array so that the odd number is located in the first half of the array, and the even number is located in the second half of the array.

Set two pointers. One is initialized to ou = 0, and the other is initialized to ji = 1. If arr [ou] is an even number, ou + = 2. If arr [ji] is an odd number, ji + = 2;

If ou <len & ji <len, then it is judged that if the even number is an odd number and the odd number is an even number, the element is exchanged.

 

Public class jiou {public static void main (String [] args) {int [] arr = {0, 1, 3, 2, 2, 4, 4, 5, 5, 5, 7 }; reorder (arr); for (int I = 0; I <arr. length; I ++) {System. out. print (arr [I] + ",") ;}} public static void reorder (int [] arr) {if (arr = null) {System. out. println ("the array is null"); return;} int len = arr. length; int ou = 0; int ji = 1; while (ou <len & ji <len) {if (arr [ou] % 2 = 0) {ou = ou + 2;} if (arr [ji] % 2 = 1) {ji = ji + 2;} if (ou <len & ji <len) {// The purpose of this condition is to first determine whether the array subscript is out of bounds. If this condition is not added, the program may have an array subscript out-of-bounds error if (arr [ou] % 2 = 1 & arr [ji] % 2 = 0) {int temp = arr [ou]; arr [ou] = arr [ji]; arr [ji] = temp ;}}}}}