• 237. Delete Node in a Linked List


    Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

    Given linked list -- head = [4,5,1,9], which looks like following:

        4 -> 5 -> 1 -> 9
    

    Example 1:

    Input: head = [4,5,1,9], node = 5
    Output: [4,1,9]
    Explanation: You are given the second node with value 5, the linked list
                 should become 4 -> 1 -> 9 after calling your function.
    

    Example 2:

    Input: head = [4,5,1,9], node = 1
    Output: [4,5,9]
    Explanation: You are given the third node with value 1, the linked list
                 should become 4 -> 5 -> 9 after calling your function.
    

    Note:

    • The linked list will have at least two elements.
    • All of the nodes' values will be unique.
    • The given node will not be the tail and it will always be a valid node of the linked list.
    • Do not return anything from your function.
      删除除了尾部外的链表中的值,一开始的做法没有考虑到释放内存,只是简单的移动赋值,奇怪的居然也AC了。
      1 void deleteNode(struct ListNode* node) {
      2     if(node){
      3         node->val=node->next->val;
      4         node->next=node->next->next;
      5     }
      6 }

      在讨论区看到别人更好的做法

      1 void deleteNode(struct ListNode* node) {
      2     if(node){
      3         struct ListNode *next=node->next;
      4         *node=*next;
      5         free(next);
      6     }
      7 }
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  • 原文地址:https://www.cnblogs.com/real1587/p/9860452.html
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