大数据面试（一）：链表问题

反转链表

public class Solution {
    public ListNode ReverseList(ListNode head) {
        ListNode dummy = new ListNode(0);
        while (head != null) {
            ListNode next = head.next;
            head.next = dummy.next;
            dummy.next = head;
            head = next;
        }
        return dummy.next;
    }
}

反转链表II

给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。

class Solution {
    public ListNode reverseBetween(ListNode head, int left, int right) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        for (int i = 0; i < left - 1; i++)
            pre = pre.next;
        ListNode cur = pre.next;
        for (int i = 0; i < right - left; i++) {
            ListNode next = cur.next;
            cur.next = next.next;
            next.next = pre.next;
            pre.next = next;
        }
        return dummy.next;
    }
}

删除链表的倒数第 K 个结点

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int k) {
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode pre = dummy;
        for (int i = 0; i < k; i++) {
            if (head == null) return null;
            head = head.next;
        }
        while (head != null) {
            pre = pre.next;
            head = head.next;
        }
        pre.next = pre.next.next;
        return dummy.next;
    }
}

K 个一组翻转链表

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode tail = head;
        for (int i = 0; i < k; i++) {
            if (tail == null) return head;
            tail = tail.next;
        }
        ListNode newHead = reverse(head, tail);
        head.next = reverseKGroup(tail, k);
        return newHead;
    }
    
    public ListNode reverse(ListNode cur, ListNode tail) {
        ListNode dummy = new ListNode(0);
        
        while (cur != tail) {
            ListNode next = cur.next;
            cur.next = dummy.next;
            dummy.next = cur;
            cur = next;
        }
        return dummy.next;
    }
}

链表是否有环？返回入环节点

public class Solution {
    public ListNode detectCycle(ListNode head) {
        if (head == null || head.next == null || head.next.next == null)
            return null;
        ListNode slow = head.next, fast = slow.next;
        while (slow != fast) {
            if (fast.next == null || fast.next.next == null)
                return null;
            slow = slow.next;
            fast = fast.next.next;
        }
        slow = head;
        while (slow != fast) {
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
    }
}

合并两个有序链表

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                cur.next = l1;  
                l1 = l1.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        if (l1 != null) {
            cur.next = l1;
        }
        if (l2 != null) {
            cur.next = l2;
        }
        return dummy.next;
    }
}

合并K个升序链表

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        return merge(lists, 0, lists.length - 1);
    }
    public ListNode merge(ListNode[] lists, int l, int r) {
        if (l == r) return lists[l];
        if (l > r) return null;
        int mid = (l + r) >> 1;
        return mergeTwoLists(merge(lists, l, mid), merge(lists, mid + 1, r));
    }
    public ListNode mergeTwoLists(ListNode a, ListNode b) {
        if (a == null) return b;
        if (b == null) return a;
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        while (a != null && b != null) {
            if (a.val < b.val) {
                cur.next = a;
                a = a.next;
            } else {
                cur.next = b;
                b = b.next;
            }
            cur = cur.next;
        }
        if (a != null) cur.next = a;
        if (b != null) cur.next = b;
        return dummy.next;
    }
}

删除排序链表中的重复元素

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) return null;
        ListNode cur = head;
        while (cur.next != null) {
            if (cur.val == cur.next.val) {
                cur.next = cur.next.next;
            } else {
                cur = cur.next;
            }
        }
        return head;
    }
}