203. Remove Linked List Elements【easy】
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.
解法一：
/**
* 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) {
if (head == NULL) {
return head;
} ListNode * dummy = new ListNode(INT_MIN);
dummy->next = head;
head = dummy; while (head != NULL && head->next != NULL) {
if (head->next->val == val) {
while (head->next != NULL && head->next->val == val) {
ListNode * temp = head->next;
free(temp);
head->next = head->next->next;
}
}
else {
head = head->next;
}
} return dummy->next;
}
};
里面的while可以不用，因为这个题和（82. Remove Duplicates from Sorted List II）不一样，那个题你根本不知道val是什么，所以只用一个if是不行的。但是这个题你知道val是什么，所以可以省略里面的while。
解法二：
/**
* 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) {
if (head == NULL) {
return head;
} ListNode * dummy = new ListNode(INT_MIN);
dummy->next = head;
head = dummy; while (head != NULL && head->next != NULL) {
if (head->next->val == val) {
ListNode * temp = head->next;
free(temp);
head->next = head->next->next;
}
else {
head = head->next;
}
} return dummy->next;
}
};
精简写法
解法三：
public ListNode removeElements(ListNode head, int val) {
if (head == null) return null;
head.next = removeElements(head.next, val);
return head.val == val ? head.next : head;
}
递归，参考了@renzid 的代码