单链表删除节点的方法

public class ListNode {
    int val;
    ListNode next;
    ListNode(int x) { val = x; }
}

删除一个单链表里的某个指定的节点:

1.修改指针指向的对象

public static void deleteNodeV2(ListNode head, ListNode node) {
    if(head == null || node == null) {
        return;
    }
    while (head != null) {
        if(head.next.val == node.val) {
            head.next = head.next.next;
            return;
        }
        head = head.next;
    }
}

2.指针指向的对象不变,节点的值覆盖,需要被删除node不是尾节点

public static void deleteNode(ListNode node) {
    if(node == null || node.next == null) {
        return;
    }
    node.val = node.next.val;
    node.next = node.next.next;
    return;
}

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转载自blog.csdn.net/Wengzhengcun/article/details/87971694