代码随想录算法训练营第三天|Leetcode203. 移除链表元素、707. 设计链表、206. 反转链表

今天的题都是链表相关的题

203. 移除链表元素

这道题还是比较简单的，用双指针来做

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode dummy = new ListNode(-1);
        ListNode cur = head;
        ListNode pre = dummy;
        dummy.next = cur;
        while(cur != null) {
            if(cur.val == val) {
                pre.next = cur.next;
            } else {
                pre = cur;
            }
            
            cur = cur.next;
        }
        return dummy.next;
    }
}

707. 设计链表

这道题之前没做过，是模拟题，我是用单链表的思路来解决的，要注意index是从0开始的

class MyLinkedList {
    class ListNode {
        int val;
        ListNode next;
        ListNode(){}
        ListNode(int _val) {
            this.val = _val;
        }
    }
    int size;
    //虚拟头节点
    ListNode dummyHead;
    public MyLinkedList() {
        this.size = 0;
        dummyHead = new ListNode(-1);
    }
    
    public int get(int index) {
        if(index < 0 || index >= size) {
            return -1;
        }
        ListNode cur = dummyHead;
        for(int i = 0; i <= index; i++) {
            cur = cur.next;
        }
        return cur.val;
    }
    
    public void addAtHead(int val) {
        addAtIndex(0, val);
    }
    
    public void addAtTail(int val) {
        addAtIndex(size, val);
    }
    
    public void addAtIndex(int index, int val) {
        if(index > size) {
            return;
        }
        if(index < 0) {
            index = 0;
        }
        size++;
        ListNode node = new ListNode(val);

        ListNode cur = dummyHead;
        for(int i = 0; i < index; i++) {
            cur = cur.next;
        }
        node.next = cur.next;
        cur.next = node;
    }
    
    public void deleteAtIndex(int index) {
        if(index >= size || index < 0) {
            return;
        }
        
        size--;
        if(index == 0) {
            dummyHead = dummyHead.next;
            return;
        }
        ListNode cur = dummyHead;
        for(int i = 0; i < index; i++) {
            cur = cur.next;
        }
        cur.next = cur.next.next;
    }
}

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * MyLinkedList obj = new MyLinkedList();
 * int param_1 = obj.get(index);
 * obj.addAtHead(val);
 * obj.addAtTail(val);
 * obj.addAtIndex(index,val);
 * obj.deleteAtIndex(index);
 */

206. 反转链表

这道题可以用迭代，也可以用递归

递归也是用迭代的思路去改写

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        return recur(head, null);
    }
    private ListNode recur(ListNode cur, ListNode pre) {
        if(cur == null)return pre;
        ListNode res = recur(cur.next, cur);
        cur.next = pre;
        return res;
    }
}