leetcode-面试题 02.04. 分割链表

leetcode原题

解题思路

只需维护两个链表 small 和 large 即可small 链表按顺序存储所有小于 xx 的节点，large 链表按顺序存储所有大于等于 xx 的节点。遍历完原链表后，我们只要将 small 链表尾节点指向 large 链表的头节点即能完成对链表的分割。

/**
 * @param {ListNode} head
 * @param {number} x
 * @return {ListNode}
 */
var partition = function(head, x) {
    let small = new ListNode(0);
    const smallHead = small;
    let large = new ListNode(0);
    const largeHead = large;
    while (head !== null) {
        if (head.val < x) {
            small.next = head;
            small = small.next;
        } else {
            large.next = head;
            large = large.next;
        }
        head = head.next;
    }
    large.next = null;
    small.next = largeHead.next;
    return smallHead.next;
};