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
這道題維持兩個可以移動的指針prev, curt. 分別初始化為prev = dummy, curt = head. 所以一開始就有prev.next = curt. 然后用curt遍歷鏈表，遇到curt.val == val的時候，就讓prev.next = curt.next來刪掉該節(jié)點梗夸；沒有遇到就直接前移prev. 兩種情況都會前移curt = curt.next來繼續(xù)遍歷。最后返回固定節(jié)點dummy.next來返回結果界轩。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode removeElements(ListNode head, int val) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode prev = dummy;
ListNode curt = head;
while (curt != null){
if (curt.val == val){
prev.next = curt.next;
} else {
prev = curt;
}
curt = curt.next;
}
return dummy.next;
}
}
