[offer18] delete the node of the linked list

problem

Given the head pointer of one-way linked list and the value of a node to be deleted , Define a function to delete the node .

Return the head node of the deleted linked list .

Be careful ： This question is different from the original one

Example 1:

Input : head = [4,5,1,9], val = 5
Output : [4,1,9]
explain : Given that the value of your list is 5 Second node of , So after calling your function , The list should be 4 -> 1 -> 9.
Example 2:

Input : head = [4,5,1,9], val = 1
Output : [4,5,9]
explain : Given that the value of your list is 1 The third node of , So after calling your function , The list should be 4 -> 5 -> 9.

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/shan-chu-lian-biao-de-jie-dian-lcof
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

class Solution {
    
public:
    /* Return the head pointer of the linked list , Or use the header node ->next Realization , Or dynamically maintain the header pointer . The code selects the second */
	ListNode* deleteNode(ListNode* head, int val) {
    
		/* Traverse a pointer of the linked list */
		ListNode* current = head;
		/* The essence of deleting a node is to find its precursor .*/
		ListNode* prior = head;
		/* The length of the linked list is not specified . Then keep cycling .*/
		while (1) {
    
			/* When the node is found */
			if (current->val == val) {
    
				/* Come up and directly find the target , The first node has no precursor . Require special treatment */
				if (current == head) {
    
				/* First consciousness is actually head++, Abandoned */
					head = head->next;
					return head;
				}
				else {
    
					prior->next = current->next;
					return head;
				}
			}
			/* If not found , Back up the current node */
			prior = current;
			current = current->next;
		}
	}
};