LeetCode（剑指 Offer）- 18. 删除链表的节点

题目链接：点击打开链接

题目大意：略

解题思路：略

相关企业

• 亚马逊（中国）投资有限公司

AC 代码

• Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
// 解决方案(1)
class Solution {
    public ListNode deleteNode(ListNode head, int val) {
        ListNode tmp = head, pre = head;
        if (head != null && head.val == val) {
            return head.next;
        }
        while (tmp != null) {
            if (tmp.val != val) {
                pre = tmp;
                tmp = tmp.next;
                continue;
            }
            pre.next = tmp.next;
            break;
        }
        return head;
    }
}

// 解决方案(2)
class Solution {
    public ListNode deleteNode(ListNode head, int val) {
        if(head.val == val) return head.next;
        ListNode pre = head, cur = head.next;
        while(cur != null && cur.val != val) {
            pre = cur;
            cur = cur.next;
        }
        if(cur != null) pre.next = cur.next;
        return head;
    }
}

• C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteNode(ListNode* head, int val) {
        if(head->val == val) return head->next;
        ListNode *pre = head, *cur = head->next;
        while(cur != nullptr && cur->val != val) {
            pre = cur;
            cur = cur->next;
        }
        if(cur != nullptr) pre->next = cur->next;
        return head;
    }
};