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java如何调整链表

2026-03-21 21:13:12Java

调整链表的方法

在Java中调整链表通常涉及插入、删除、修改节点或重新组织链表结构。以下是常见的操作示例：

插入节点

在链表头部或尾部插入新节点：

java如何调整链表

// 头部插入
public void insertAtHead(ListNode head, int val) {
    ListNode newNode = new ListNode(val);
    newNode.next = head;
    head = newNode;
}

// 尾部插入
public void insertAtTail(ListNode head, int val) {
    ListNode newNode = new ListNode(val);
    if (head == null) {
        head = newNode;
        return;
    }
    ListNode curr = head;
    while (curr.next != null) {
        curr = curr.next;
    }
    curr.next = newNode;
}

删除节点

删除指定值的节点：

public ListNode deleteNode(ListNode head, int val) {
    if (head == null) return null;
    if (head.val == val) return head.next;

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

反转链表

反转整个链表：

java如何调整链表

public ListNode reverseList(ListNode head) {
    ListNode prev = null;
    ListNode curr = head;
    while (curr != null) {
        ListNode nextTemp = curr.next;
        curr.next = prev;
        prev = curr;
        curr = nextTemp;
    }
    return prev;
}

交换相邻节点

交换链表中相邻的两个节点：

public ListNode swapPairs(ListNode head) {
    if (head == null || head.next == null) return head;

    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode curr = dummy;

    while (curr.next != null && curr.next.next != null) {
        ListNode first = curr.next;
        ListNode second = curr.next.next;

        first.next = second.next;
        curr.next = second;
        curr.next.next = first;
        curr = curr.next.next;
    }
    return dummy.next;
}

合并两个有序链表

将两个有序链表合并为一个：

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode curr = dummy;

    while (l1 != null && l2 != null) {
        if (l1.val < l2.val) {
            curr.next = l1;
            l1 = l1.next;
        } else {
            curr.next = l2;
            l2 = l2.next;
        }
        curr = curr.next;
    }

    curr.next = (l1 != null) ? l1 : l2;
    return dummy.next;
}