9, the order of the array so that the adjustment in front of even-odd ------------> Series offer to prove safety

Title Description

Enter an array of integers, to realize a function to adjust the order of the numbers in the array, such that all the odd part of the front half of the array, is located in the second half of all the even array, and between odd and ensure a relatively even, odd and even the same position.

Thinking

First, find the first odd, removed, and then the odd element before the first all moved back one, first find the odd talk on the 0 position.

Looking at the odd element sequentially after the first odd number, and the moving operation done. We can ensure that the original relative order.

function reOrderArray(array)
{
    let j=0;
    let m=0;
    for(let i=0;i<array.length;i++){
        if(array[i]%2==1){
            let temp=array[i];
            let ti=i;
            for (; ti> 0; T i -) {
                array [ti] = array [ti-1]
            }
            array[0]=temp;
            j=i;
            break;
        }
    }
    for(++j;j<array.length;j++){
        if(array[j]%2==1){
            let temp=array[j];
            let tj=j;
            for(;tj>m;tj--){
                array[tj]=array[tj-1];
            }
            array[++m]=temp;
        }
    }
    return array;
}

If you do not consider the relative position is:

Thinking

Setting two pointers

The first pointer to start from the first element of the array, forward to aft

The second pointer from the end of the last element in the array, advancing to the head

When traversing to even start, end traversed odd, exchange the position of two numbers

When the start> end, the complete exchange

Code

    function reOrderArray(array) {
      if (Array.isArray(array)) { let start = 0; let end = array.length - 1; while (start < end) { while (array[start] % 2 === 1) { start++; } while (array[end] % 2 === 0) { end--; } if (start < end) { [array[start], array[end]] = [array[end], array[start]] } } } return array; }