题解 | #删除链表的节点#

/*
 * 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
 * @param val int整型
 * @return ListNode类
 */
function deleteNode(head, val) {
    let cur = head;
    let pre = null;
    // 特殊情况：删除头节点
    if(head.val === val){
        head = head.next;
        return head;
    }
    // 其余情况
    while (cur) {
        if (cur.val === val) {
            pre.next = cur.next;
            return head;
        }
        pre = cur;
        cur = cur.next;
    }
}
module.exports = {
    deleteNode: deleteNode,
};