题解 | #链表的奇偶重排#

/*
 * function ListNode(x){
 *   this.val = x;
 *   this.next = null;
 * }
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param head ListNode类 
 * @return ListNode类
 */
function oddEvenList( head ) {
    // write code here
    let p1 = head
    let current = 1
    let ji = new ListNode(0)
    let currentJi = ji
    let ou = new ListNode(0)
    let currentOu = ou
    while(p1) {
        if (current % 2) { //奇
            currentJi.next = p1
            currentJi = currentJi.next
        }
        else {
            currentOu.next = p1
            currentOu = currentOu.next
        }
        p1 = p1.next
        current++
    }
    currentOu.next = null
    currentJi.next = ou.next
    return ji.next
}
module.exports = {
    oddEvenList : oddEvenList
};