[LeetCode]84. Remove Linked List Elements移除链表元素

Remove all elements from a linked list of integers that have value val.

Example
Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
Return: 1 --> 2 --> 3 --> 4 --> 5

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

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解法1：（1）如果链表最前面的若干个元素就是待删除元素，则删除之；（2）如果此时链表非空，则定义两个指针curr和next指向相邻两个节点，并不断往后遍历。如果next指向节点待删除节点，则删除之。直到next元素为尾节点。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        while(head != NULL && head->val == val) {
            ListNode* del = head;
            head = head->next;
            delete del;
        }
        if(head == NULL) return NULL;
        ListNode *curr = head, *next = head->next;
        while(next != NULL) {
            if(next->val == val) {
                ListNode* del = next;
                next = next->next;
                curr->next = next;
                delete del;
            }
            else {
                curr = next;
                next = next->next;
            }
        }
        return head;
    }
};

注意next节点为待删除节点和非删除节点时处理的差异。

解法2：更方便的方法是在头节点前加入一个辅助节点，那么处理头节点就可以和处理普通节点一样了。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode* help = new ListNode(0);
        help->next = head;
        ListNode *curr = help, *next = head;
        while(next != NULL) {
            if(next->val == val) {
                ListNode* del = next;
                next = next->next;
                curr->next = next;
                delete del;
            }
            else {
                curr = next;
                next = next->next;
            }
        }
        return help->next;
    }
};