[LeetCode]Delete Node in a Linked List

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

Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after calling your function.

［思路］只知道当前节点，要删除当前节点。和通常的删除不同的是，我们不知道前一个节点。

我们可以利用当前节点与前一个节点的联系，把后一个节点拷贝过来。然后删除后一个节点，相当于实现了删除操作。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void deleteNode(ListNode* node) {
        if(node == nullptr)
            return;
        ListNode *next = node->next;
        node->val = node->next->val;
        node->next = next->next;
        delete next;
    }
};

[LeetCode]Delete Node in a Linked List

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原文地址：http://blog.csdn.net/ciaoliang/article/details/46939937