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
这道题维持两个可以移动的指针prev, curt. 分别初始化为prev = dummy, curt = head. 所以一开始就有prev.next = curt. 然后用curt遍历链表，遇到curt.val == val的时候，就让prev.next = curt.next来删掉该节点；没有遇到就直接前移prev. 两种情况都会前移curt = curt.next来继续遍历。最后返回固定节点dummy.next来返回结果。
/**
* 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;
}
}
