Leetcode 题解 链表

2019/05/18 CyC

文章来源:https://github.com/CyC2018/CS-Notes

链表是空节点,或者有一个值和一个指向下一个链表的指针,因此很多链表问题可以用递归来处理。

1. 找出两个链表的交点

160. Intersection of Two Linked Lists (Easy)

例如以下示例中 A 和 B 两个链表相交于 c1:

A:          a1 → a2
                    ↘
                      c1 → c2 → c3
                    ↗
B:    b1 → b2 → b3

但是不会出现以下相交的情况,因为每个节点只有一个 next 指针,也就只能有一个后继节点,而以下示例中节点 c 有两个后继节点。

A:          a1 → a2       d1 → d2
                    ↘  ↗
                      c
                    ↗  ↘
B:    b1 → b2 → b3        e1 → e2

要求时间复杂度为 O(N),空间复杂度为 O(1)。如果不存在交点则返回 null。

设 A 的长度为 a + c,B 的长度为 b + c,其中 c 为尾部公共部分长度,可知 a + c + b = b + c + a。

当访问 A 链表的指针访问到链表尾部时,令它从链表 B 的头部开始访问链表 B;同样地,当访问 B 链表的指针访问到链表尾部时,令它从链表 A 的头部开始访问链表 A。这样就能控制访问 A 和 B 两个链表的指针能同时访问到交点。

如果不存在交点,那么 a + b = b + a,以下实现代码中 l1 和 l2 会同时为 null,从而退出循环。

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    ListNode l1 = headA, l2 = headB;
    while (l1 != l2) {
        l1 = (l1 == null) ? headB : l1.next;
        l2 = (l2 == null) ? headA : l2.next;
    }
    return l1;
}

如果只是判断是否存在交点,那么就是另一个问题,即 编程之美 3.6 的问题。有两种解法:

  • 把第一个链表的结尾连接到第二个链表的开头,看第二个链表是否存在环;
  • 或者直接比较两个链表的最后一个节点是否相同。

2. 链表反转

206. Reverse Linked List (Easy)

递归

public ListNode reverseList(ListNode head) {
    if (head == null || head.next == null) {
        return head;
    }
    ListNode next = head.next;
    ListNode newHead = reverseList(next);
    next.next = head;
    head.next = null;
    return newHead;
}

头插法

public ListNode reverseList(ListNode head) {
    ListNode newHead = new ListNode(-1);
    while (head != null) {
        ListNode next = head.next;
        head.next = newHead.next;
        newHead.next = head;
        head = next;
    }
    return newHead.next;
}

3. 归并两个有序的链表

21. Merge Two Sorted Lists (Easy)

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

4. 从有序链表中删除重复节点

83. Remove Duplicates from Sorted List (Easy)

Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.
public ListNode deleteDuplicates(ListNode head) {
    if (head == null || head.next == null) return head;
    head.next = deleteDuplicates(head.next);
    return head.val == head.next.val ? head.next : head;
}

5. 删除链表的倒数第 n 个节点

19. Remove Nth Node From End of List (Medium)

Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode fast = head;
    while (n-- > 0) {
        fast = fast.next;
    }
    if (fast == null) return head.next;
    ListNode slow = head;
    while (fast.next != null) {
        fast = fast.next;
        slow = slow.next;
    }
    slow.next = slow.next.next;
    return head;
}

6. 交换链表中的相邻结点

24. Swap Nodes in Pairs (Medium)

Given 1->2->3->4, you should return the list as 2->1->4->3.

题目要求:不能修改结点的 val 值,O(1) 空间复杂度。

public ListNode swapPairs(ListNode head) {
    ListNode node = new ListNode(-1);
    node.next = head;
    ListNode pre = node;
    while (pre.next != null && pre.next.next != null) {
        ListNode l1 = pre.next, l2 = pre.next.next;
        ListNode next = l2.next;
        l1.next = next;
        l2.next = l1;
        pre.next = l2;

        pre = l1;
    }
    return node.next;
}

7. 链表求和

445. Add Two Numbers II (Medium)

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

题目要求:不能修改原始链表。

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    Stack<Integer> l1Stack = buildStack(l1);
    Stack<Integer> l2Stack = buildStack(l2);
    ListNode head = new ListNode(-1);
    int carry = 0;
    while (!l1Stack.isEmpty() || !l2Stack.isEmpty() || carry != 0) {
        int x = l1Stack.isEmpty() ? 0 : l1Stack.pop();
        int y = l2Stack.isEmpty() ? 0 : l2Stack.pop();
        int sum = x + y + carry;
        ListNode node = new ListNode(sum % 10);
        node.next = head.next;
        head.next = node;
        carry = sum / 10;
    }
    return head.next;
}

private Stack<Integer> buildStack(ListNode l) {
    Stack<Integer> stack = new Stack<>();
    while (l != null) {
        stack.push(l.val);
        l = l.next;
    }
    return stack;
}

8. 回文链表

234. Palindrome Linked List (Easy)

题目要求:以 O(1) 的空间复杂度来求解。

切成两半,把后半段反转,然后比较两半是否相等。

public boolean isPalindrome(ListNode head) {
    if (head == null || head.next == null) return true;
    ListNode slow = head, fast = head.next;
    while (fast != null && fast.next != null) {
        slow = slow.next;
        fast = fast.next.next;
    }
    if (fast != null) slow = slow.next;  // 偶数节点,让 slow 指向下一个节点
    cut(head, slow);                     // 切成两个链表
    return isEqual(head, reverse(slow));
}

private void cut(ListNode head, ListNode cutNode) {
    while (head.next != cutNode) {
        head = head.next;
    }
    head.next = null;
}

private ListNode reverse(ListNode head) {
    ListNode newHead = null;
    while (head != null) {
        ListNode nextNode = head.next;
        head.next = newHead;
        newHead = head;
        head = nextNode;
    }
    return newHead;
}

private boolean isEqual(ListNode l1, ListNode l2) {
    while (l1 != null && l2 != null) {
        if (l1.val != l2.val) return false;
        l1 = l1.next;
        l2 = l2.next;
    }
    return true;
}

9. 分隔链表

725. Split Linked List in Parts(Medium)

Input:
root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
Explanation:
The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.

题目描述:把链表分隔成 k 部分,每部分的长度都应该尽可能相同,排在前面的长度应该大于等于后面的。

public ListNode[] splitListToParts(ListNode root, int k) {
    int N = 0;
    ListNode cur = root;
    while (cur != null) {
        N++;
        cur = cur.next;
    }
    int mod = N % k;
    int size = N / k;
    ListNode[] ret = new ListNode[k];
    cur = root;
    for (int i = 0; cur != null && i < k; i++) {
        ret[i] = cur;
        int curSize = size + (mod-- > 0 ? 1 : 0);
        for (int j = 0; j < curSize - 1; j++) {
            cur = cur.next;
        }
        ListNode next = cur.next;
        cur.next = null;
        cur = next;
    }
    return ret;
}

10. 链表元素按奇偶聚集

328. Odd Even Linked List (Medium)

Example:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.
public ListNode oddEvenList(ListNode head) {
    if (head == null) {
        return head;
    }
    ListNode odd = head, even = head.next, evenHead = even;
    while (even != null && even.next != null) {
        odd.next = odd.next.next;
        odd = odd.next;
        even.next = even.next.next;
        even = even.next;
    }
    odd.next = evenHead;
    return head;
}

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