Linked List Mock

weixin_652790062024-04-09 1:30

203. Remove Linked List Elements

Solved

Easy

Topics

Companies

Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.

java 复制代码 
/**
 * 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 dummyHead = new ListNode(-1,head);
        ListNode cur = dummyHead;
        while(cur != null && cur.next != null){
            if(cur.next.val == val){
                cur.next = cur.next.next;
            }else{
                cur = cur.next;
            }
        }
        return dummyHead.next;
    }
}

206. Reverse Linked List

Solved

Easy

Topics

Companies

Given the head of a singly linked list, reverse the list, and return the reversed list.

java 复制代码 
/**
 * 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) {
        ListNode newHead = null;
        ListNode o1 = head;
        while(o1 != null){
            ListNode o2 = o1.next;
            o1.next = newHead;
            newHead = o1;
            o1 = o2;
        }
        return newHead;
    }
}
java 复制代码 
/**
 * 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) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode last = reverseList(head.next);
        head.next.next = head;
        head.next = null;
        return last;
    }
}

24. Swap Nodes in Pairs

Solved

Medium

Topics

Companies

Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)

java 复制代码 
/**
 * 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 swapPairs(ListNode head) {
        ListNode dummyHead = new ListNode(-1 , head);
        ListNode cur = dummyHead;
        ListNode n1 = head;
        while(n1 != null && n1.next != null){
            ListNode n2 = n1.next.next;
            cur.next = n1.next;
            n1.next.next = n1;
            n1.next = n2;
            cur = n1;
            n1 = n2;
        }
        return dummyHead.next;
    }
}
java 复制代码 
/**
 * 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 swapPairs(ListNode head) {
        if(head == null||head.next==null){
            return head;
        }
        ListNode newHead = head.next;
        head.next = swapPairs(newHead.next);
        newHead.next = head;
        return newHead;
    }
}

19. Remove Nth Node From End of List

Solved

Medium

Topics

Companies

Hint

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

java 复制代码 
/**
 * 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 removeNthFromEnd(ListNode head, int n) {
        ListNode dummyHead = new ListNode(-1, head);
        ListNode n1 = dummyHead;
        ListNode n2 = dummyHead;
        for(int i = 0; i < n + 1; i++){
            n2 = n2.next;
        }
        while(n2 != null){
            n2 = n2.next;
            n1 = n1.next;
        }
        n1.next = n1.next.next;
        return dummyHead.next;
    }
}

141. Linked List Cycle

Easy

Given head, the head of a linked list, determine if the linked list has a cycle in it.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter.

Return trueif there is a cycle in the linked list . Otherwise, return false.

java 复制代码 
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null) {
            return false;
        }
        ListNode slow = head;
        ListNode fast = head.next;
        while(fast!=null && fast.next!=null){
            if(slow == fast) return true;
            slow=slow.next;
            fast=fast.next.next;
        }
        return false;
    }
}

142. Linked List Cycle II

Solved

Medium

Topics

Companies

Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to (0-indexed ). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

Do not modify the linked list.

java 复制代码 
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
         if(head==null){
            return null;
        }
        ListNode fast = head;
        ListNode slow  =head;
        while(fast!=null && fast.next!=null){
            fast = fast.next.next;
            slow = slow.next;
            if(slow == fast){
                slow = head;
                while(slow!=fast){
                    slow = slow.next;
                    fast = fast.next;
                }
                return slow;
            }
        }
        return null;
    }
}

21. Merge Two Sorted Lists

Solved

Easy

Topics

Companies

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

java 复制代码 
/**
 * 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 mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode dummyHead = new ListNode(-1,null);
        ListNode n = dummyHead;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                n.next = list1;
                list1 = list1.next;
            } else {
                n.next = list2;
                list2 = list2.next;
            }
            n = n.next;
        }
        if (list1 == null) {
            n.next = list2;
        }
        if (list2 == null) {
            n.next = list1;
        }
        return dummyHead.next;
    }
}

82. Remove Duplicates from Sorted List II

Solved

Medium

Topics

Companies

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

java 复制代码 
/**
 * 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 deleteDuplicates(ListNode head) {
        ListNode dummyHead = new ListNode(-1 , head);
        ListNode slow = dummyHead;
        ListNode fast = head;
        while(fast != null && fast.next != null){
            if(fast.val == fast.next.val){
                int val = fast.val;
                while(fast != null && fast.val == val){
                    fast = fast.next;
                    slow.next = fast;
                }
            }else{
                fast = fast.next;
                slow = slow.next;
            }
        }
        return dummyHead.next;
    }
}
java 复制代码 
/**
 * 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 deleteDuplicates(ListNode head) {
        ListNode dummyHead = new ListNode(-1, head);
        ListNode cur = dummyHead;
        while (cur.next != null && cur.next.next != null) {
            if (cur.next.val == cur.next.next.val) {
                ListNode p = cur.next;
                while (cur.next != null && cur.next.val == p.val) {
                    cur.next = cur.next.next;
                }
            } else {
                cur = cur.next;
            }
        }
        return dummyHead.next;
    }
}

83. Remove Duplicates from Sorted List

Solved

Easy

Topics

Companies

Given the head of a sorted linked list, delete all duplicates such that each element appears only once . Return the linked list sorted as well.

java 复制代码 
/**
 * 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 deleteDuplicates(ListNode head) {
        ListNode cur = head;
        while(cur != null && cur.next != null){
            if(cur.next.val == cur.val){
                cur.next = cur.next.next;
            }else{
                cur = cur.next;
            }
        }
        return head;
    }
}

160. Intersection of Two Linked Lists

Solved

Easy

Topics

Companies

Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect . If the two linked lists have no intersection at all, return null.

java 复制代码 
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode curA = headA;
        ListNode curB = headB;
        while(curA != curB){
            if(curA != null){
                curA = curA.next;
            }else{
                curA = headB;
            }
            if(curB != null){
                curB = curB.next;
            }else{
                curB = headA;
            }
        }
        return curA;
    }
}
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