leetcode linked list

21. 合并两个有序链表

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        else if (l2 == null) {
            return l1;
        }
        else if (l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        }
        else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }

    }
}

83. 删除排序链表中的重复元素

    ListNode current = head;
    while (current != null && current.next != null) {
        if (current.next.val == current.val) {
            current.next = current.next.next;
        } else {
            current = current.next;
        }
    }
    return head;
}

21. 合并两个有序链表

    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null)
            return head;
        ListNode fast = head.next, slow = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode tmp = slow.next;
        slow.next = null;
        ListNode left = sortList(head);
        ListNode right = sortList(tmp);
        ListNode h = new ListNode(0);
        ListNode res = h;
        while (left != null && right != null) {
            if (left.val < right.val) {
                h.next = left;
                left = left.next;
            } else {
                h.next = right;
                right = right.next;
            }
            h = h.next;
        }
        h.next = left != null ? left : right;
        return res.next;
    }
}

160. 相交链表

第三个想法就是说两个节点其实是走完相同的路程了，这样有相交的节点，肯定要相遇，如果没有相遇，那么就是走到null了。。。

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) {
            return null;
        }
        ListNode pA = headA;
        ListNode pB = headB;
        while (pA != pB) {
            pA = pA.next;
            pB = pB.next;
            if (pA == null && pB == null) {
                return null;
            }
            if (pA == null) {
                pA = headB;
            }
            if (pB == null) {
                pB = headA;
            }
        }
        return pA;
    }
}

206. 反转链表

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode pre = null;
        ListNode curr = head;
        while(curr!= null) {
            ListNode next = curr.next;
            curr.next = pre;
            pre = curr;
            curr = next;
        }
        return pre;
    }
}

141. 环形链表

public boolean hasCycle(ListNode head) {
    if (head == null || head.next == null) {
        return false;
    }
    ListNode slow = head;
    ListNode fast = head.next;
    while (slow != fast) {
        if (fast == null || fast.next == null) {
            return false;
        }
        slow = slow.next;
        fast = fast.next.next;
    }
    return true;
}

19. 删除链表的倒数第N个节点

public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode first = dummy;
    ListNode second = dummy;
    // Advances first pointer so that the gap between first and second is n nodes apart
    for (int i = 1; i <= n + 1; i++) {
        first = first.next;
    }
    // Move first to the end, maintaining the gap
    while (first != null) {
        first = first.next;
        second = second.next;
    }
    second.next = second.next.next;
    return dummy.next;
}