代码随想录算法训练营第3天| 203.移除链表元素, 707.设计链表, 206.反转链表

203.移除链表元素

自己看到题目的第一想法

Very simple delete operation of linkedlist. The technique is creating a dummyHead in front of the linkedlist, so that deleting the head is the same as deleting the other node.

class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode dummy = new ListNode(-1);
        ListNode ans = dummy;
        dummy.next = head;
        while (dummy != null) {
            if (dummy.next != null && dummy.next.val == val) {
                dummy.next = dummy.next.next;
                continue;
            }
            dummy = dummy.next;
        }
        return ans.next;
    }
}

707.设计链表

自己看到题目的第一想法

This question tested my understanding of the major operations of a linkedlist. The difficult point for me is addAtIndex and deleteAtIndex. The key here is I need to find the Node before the index in order to insert or delete the Node.

class MyLinkedList {

    class ListNode {
        int val;
        ListNode next;

        public ListNode(int val) {
            this.val = val;
            this.next = null;
        }

        public ListNode(int val, ListNode next) {
            this.val = val;
            this.next = next;
        }
    }

    ListNode dummyHead;

    public MyLinkedList() {
        this.dummyHead = new ListNode(-1);
    }
    
    public int get(int index) {
        if (dummyHead.next == null) {
            return -1;
        }
        ListNode cur = dummyHead.next;
        int count = 0;
        while (cur != null) {
            if (index == count) {
                return cur.val;
            }
            count++;
            cur = cur.next;
        }
        return -1;
    }
    
    public void addAtHead(int val) {
        ListNode headNext = dummyHead.next;
        dummyHead.next = new ListNode(val, headNext);
    }
    
    public void addAtTail(int val) {
        ListNode cur = this.dummyHead;
        while (cur.next != null) {
            cur = cur.next;
        }
        cur.next = new ListNode(val);
    }
    
    public void addAtIndex(int index, int val) {
        ListNode cur = dummyHead;
        int nextIdx = 0;
        while (cur.next != null) {
            if (nextIdx == index) {
                ListNode nextnext = cur.next;
                cur.next = new ListNode(val, nextnext);
                return;
            }
            nextIdx++;
            cur = cur.next;
        }
        if (nextIdx == index) {
            cur.next = new ListNode(val);
        }
    }
    
    public void deleteAtIndex(int index) {
        int count = 0;
        ListNode cur = dummyHead;
        while (cur != null && cur.next != null) {
            if (count == index) {
                cur.next = cur.next.next;
            }
            cur = cur.next;
            count++;
        }
    }
}

206.反转链表

自己看到题目的第一想法

Just some manipulation of pointer in linkedlist, can be done in 1 pass.

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode end = null;
        ListNode cur = head;
        while (cur != null) {
            ListNode curNext = cur.next;
            cur.next = end;
            end = cur;
            cur = curNext;
        }
        return end;
    }
}