android-notes/blogs/Algorithm/剑指 Offer/链表相关.md

链表相关

目录

1. 06. 从头到尾打印链表
2. 18. 删除链表的节点
3. 22. 链表中倒数第 k 个节点
4. 24. 反转链表
5. 25. 合并两个排序的链表
6. 35. 复杂链表的复制

06. 从头到尾打印链表

class Solution {
    public int[s] reversePrint(ListNode head) {
        ListNode temp = head;
        int size = 0;
        while (temp != null) {
            temp = temp.next;
            size++;
        }
        int[] result = new int[size];
        int index = size - 1;
        while (head != null) {
            result[index--] = head.val;
            head = head.next;
        }
        return result;
    }
}

class Solution {
    public int[] reversePrint(ListNode head) {
        Stack<ListNode> stack = new Stack<ListNode>();
        ListNode temp = head;
        while (temp != null) {
            stack.push(temp);
            temp = temp.next;
        }
        int size = stack.size();
        int[] print = new int[size];
        for (int i = 0; i < size; i++) {
            print[i] = stack.pop().val;
        }
        return print;
    }
}

18. 删除链表的节点

class Solution {
    public ListNode deleteNode(ListNode head, int val) {
        ListNode h0 = new ListNode(0);
        ListNode h1 = h0;
        h0.next = head;
        while (h0.next != null) {
            if (h0.next.val == val) {
                h0.next = h0.next.next;
                break;
            }
            h0 = h0.next;
        }
        return h1.next;
    }
}

22. 链表中倒数第 k 个节点

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        int length = 0;
        ListNode temp = head;
        while (temp != null) {
            temp = temp.next;
            length++;
        }
        for (int i = 0; i < length - k; i++) {
            head = head.next;
        }
        return head;
    }
}

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        ListNode h0 = head;
        for (int i = 0; i < k; i++) {
            h0 = h0.next;
        }
        while (h0 != null) {
            h0 = h0.next;
            head = head.next;
        }
        return head;
    }
}

24. 反转链表

    class Solution {
        public ListNode reverseList(ListNode head) {
            if (head == null || head.next == null) {
                return head;
            }
            ListNode h1 = head;
            ListNode h2 = head.next;
            ListNode h3 = null;
            h1.next = null;
            while (h2 != null) {
                h3 = h2.next;
                h2.next = h1;
                h1 = h2;
                h2 = h3;
            }
            return h1;
        }
    }

    class Solution {
        public ListNode reverseList(ListNode head) {
            // 递归终止条件是当前为空，或者下一个节点为空
            if (head == null || head.next == null) {
                return head;
            }
            // 这里的 h1 就是最后一个节点
            ListNode h1 = reverseList(head.next);
            // 如果链表是 1->2->3->4->5，那么此时的 cur 就是 5
            // 而 head 是4，head的 下一个是 5，下下一个是空
            // 所以 head.next.next 就是 5->4
            head.next.next = head;
            // 防止链表循环，需要将 head.next 设置为空
            head.next = null;
            // 每层递归函数都返回 h1，也就是最后一个节点
            return h1;
        }
    }

25. 合并两个排序的链表

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode h0 = new ListNode(0);
        ListNode h = h0;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                h0.next = l1;
                l1 = l1.next;
            } else {
                h0.next = l2;
                l2 = l2.next;
            }
            h0 = h0.next;
        }
        if (l1 == null) {
            h0.next = l2;
        } else {
            h0.next = l1;
        }
        return h.next;
    }
}

class Solution {
    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;
        }
    }
}

35. 复杂链表的复制

class Solution {
    public Node copyRandomList(Node head) {
        // 存原节点和拷贝节点的一个映射
        Map<Node, Node> map = new HashMap<>();
        Node temp = head;
        while (temp != null) {
            map.put(temp, new Node(temp.val));
            temp = temp.next;
        }
        temp = head;
        while (temp != null) {
            // 将拷贝节点组织成一个链表
            map.get(temp).random = map.get(temp.random);
            map.get(temp).next = map.get(temp.next);
            temp = temp.next;
        }
        return map.get(head);
    }
}

class Solution {
    public Node copyRandomList(Node head) {
        // 将拷贝节点放在原节点后面，比如 1->2 变成 1->1'->2->2'
        for (Node node = head, copy = null; node != null; node = node.next.next) {
            copy = new Node(node.val);
            copy.next = node.next;
            node.next = copy;
        }
        // 把拷贝节点的 random 指针安排上
        for (Node node = head; node != null; node = node.next.next) {
            if (node.random != null) {
                node.next.random = node.random.next;
            }
        }
        Node newHead = head.next;
        Node node = head;
        Node temp = null;
        // 分离原节点和拷贝节点
        while (node != null && node.next != null) {
            temp = node.next;
            node.next = temp.next;
            node = temp;
        }
        return newHead;
    }
}

52. 两个链表的第一个公共节点

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode h1 = headA;
        ListNode h2 = headB;
        while (h1 != null && h2 != null) {
            if (h1 == h2) {
                return h1;
            }
            h1 = h1.next;
            h2 = h2.next;
            if (h1 == null && h2 == null) {
                return null;
            }
            if (h1 == null) {
                h1 = headB;
            }
            if (h2 == null) {
                h2 = headA;
            }
        }
        return null;
    }
}

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode h1 = headA;
        ListNode h2 = headB;
        while (h1 != h2) {
            h1 = h1 == null ? headB : h1.next;
            h2 = h2 == null ? headA : h2.next;
        }
        return h1;
    }
}