The sword refers to the offer------(sort, quick idea)---Adjust the array order so that the odd digits precede the even number

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https://www.nowcoder.com/practice/beb5aa231adc45b2a5dcc5b62c93f593?tpId=13&tqId=11166&rp=1&ru=% 2fta%2fcoding-interviews&qru=%2fta%2fcoding-interviews%2fquestion-ranking&tpage=1   Test Instructions title: Enter an array of integers to implement a function to adjust the order of the numbers in the array so that all the odd digits are in the first half of the array, and all the even digits are in the second half of the array. The relative position does not change.   Analysis http://blog.csdn.net/lks1139230294/article/details/52761479 1: Because the even number is put on the last side, so an even number appears at the end, Moving the rest of the part forward at the same time will ensure that the relative position does not change. O (n^2) Idea 2: First to find an even, even to the back of it, so from the even start to find the odd number, found the odd number ahead of the even position, because the odd number found is the first odd number after the even, so the odd and even between the two must be moved after one.    Code: public class Solution {    public void Reorderarray (int [] array) {         if (array = = NULL | | array.length <= 1) return;     & Nbsp;          int first = 0;         int second = 0;       //This side why use while without for, because it is not step by step, but can be directly to the end. The while is used for continuous post-move operations.       &Nbsp; while (First < Array.Length) {             while (First < Array.Length && isodd (Array[first])) {                 first ++;             }            if (First >= ARRAY.LENGTH-1) {                 break;            }             second = First + 1;             while (Second < Array.Length &&!isodd (Array[second])) {                 second ++;             }             if (Second > Array.length-1) {                 break;            }             int temp = array[second];             for (int i = second-1; I >= first; i--) {&NBSP;&NBSP;&NB Sp;             array[i + 1] = array[i];             }             array[first] = temp;        }     }        public boolean isodd (int num) {  &nbsP;     return (num & 1) = = 1;    }}

Sword refers to offer------(sort, quick idea)---Adjust the array order so that the odd digits precede the even number