题解 | #单链表的排序#

/*
 * 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 the head node
 * @return ListNode类
 */
function sortInList( head ) {
    // write code here
    if (head.next === null) {
        return head
    }
    let slow = head
    let fast = head
    let prev = null
    while (fast && fast.next) {
        prev = slow
        slow = slow.next
        fast = fast.next.next
    }
    prev.next = null
    let left = sortInList(head)
    let right = sortInList(slow)
    return mergeNode(left,right)
}

function mergeNode( head1 , head2 ) {
    let p = new ListNode(0)
    let p1 = p
    while (head1 && head2) {
        if(head1.val <= head2.val) {
            p.next = head1
            head1 = head1.next
        }
        else {
            p.next = head2
            head2 = head2.next
        }
        p = p.next
    }
    if (head1) {
        p.next = head1
    }
    if (head2) {
        p.next = head2
    }
    
    return p1.next
}
module.exports = {
    sortInList : sortInList
};