LeetCode删链表节点系列总结LeetCode 83. Remove Duplicates from Sorted List;

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本文介绍了四种链表操作的方法，包括去除重复元素、移除倒数第N个节点及移除特定值的所有节点等。通过具体实例展示了每种操作的实现过程。

83. Remove Duplicates from Sorted List

Description Hints Submissions Solutions

Total Accepted: 175237
Total Submissions: 444457
Difficulty: Easy
Contributor: LeetCode

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

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public ListNode deleteDuplicates(ListNode head) {
    if(head==null) return head;
    ListNode pre=head;
    ListNode cur=head.next;
    while(cur!=null){
        if(cur.val==pre.val){
            pre.next=cur.next;
            cur=cur.next;
            //不用修改pre
        }else{
            pre=pre.next;
            cur=cur.next;
        }
    }
    return head;
}

82. Remove Duplicates from Sorted List II

Description Hints Submissions Solutions

Total Accepted: 104206
Total Submissions: 358219
Difficulty: Medium
Contributor: LeetCode

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

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public ListNode deleteDuplicates(ListNode head) {
    ListNode helper = new ListNode(0);
    ListNode pre = helper;
    boolean isDup = false;
    while (head != null && head.next != null) {
        if (head.val == head.next.val) {
            isDup = true;
        } else if (isDup ) {
            isDup = false;
        } else {
            pre.next = head;
            pre = pre.next;
        }
        head = head.next;
    }
    if (isDup) {
        pre.next = null;
    } else {
        pre.next = head;
    }
    return helper.next;
}

19. Remove Nth Node From End of List

Description Hints Submissions Solutions

Total Accepted: 170940
Total Submissions: 520559
Difficulty: Medium
Contributor: LeetCode

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

题解：首先先让faster从起始点往后走n步。

然后再让slower和faster一起走，直到faster.next==null时候，slower所指向的node就是需要删除的节点的前一个节点。

public ListNode removeNthFromEnd(ListNode head, int n){
    ListNode fast=head;
    ListNode slow=head;
    while(n>0){
        fast=fast.next;
        n=n-1;
    }
    //要删除的是头结点
    if(fast==null){
        head=head.next;
        return head;
    }
    //没有用fast!=null,使得slow取代了pre的作用
    while (fast.next!=null){
        slow=slow.next;
        fast=fast.next;
    }

    slow.next=slow.next.next;
    return head;
}

203. Remove Linked List Elements

Description Hints Submissions Solutions

Total Accepted: 108962
Total Submissions: 342991
Difficulty: Easy
Contributor: LeetCode

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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public ListNode removeElementsI(ListNode head, int val) {
    if(head==null) return head;
    head.next=removeElementsI(head.next, val);

    return head.val==val?head.next:head;
}
public ListNode removeElements(ListNode head, int val) {
    if(head==null) return head;
    ListNode pre=new ListNode(0);
    pre.next=head;
    ListNode res= pre;
    while(head!=null){
        if(head.val==val){
            pre.next=head.next;
        }else {
            pre=pre.next;
        }
        head=head.next;
    }
    return res.next;
}