《Cracking the Coding Interview》——第2章：链表——题目3

2014-03-18 02:25

题目：给定一个单链表中间的节点，删掉那个节点。

解法：把后面节点的数据域拷到当前节点来，然后删除后面那个节点。当前节点不是尾巴，所以后面不为空。

代码：

 // 2.2 Remove a node from middle of a linked list
 #include <cstdio>
 using namespace std;

 struct ListNode {
     int val;
     struct ListNode *next;
     ListNode(int x): val(x), next(nullptr) {};
 };

 class Solution {
 public:
     void deleteThatNode(ListNode *ptr) {
         if (ptr != nullptr && ptr->next != nullptr) {
             struct ListNode *tmp = ptr->next;
             ptr->val = tmp->val;
             ptr->next = tmp->next;
             delete tmp;
         }
     }
 };

 int main()
 {
     int i;
     int n, k;
     int val;
     struct ListNode *head, *ptr;
     Solution sol;

     while (scanf("%d", &n) ==  && n > ) {
         // create a linked list
         ptr = head = nullptr;
         for (i = ; i < n; ++i) {
             scanf("%d", &val);
             if (head == nullptr) {
                 head = ptr = new ListNode(val);
             } else {
                 ptr->next = new ListNode(val);
                 ptr = ptr->next;
             }
         }

         // remove a node from middle of the list
         scanf("%d", &k);
         k = k <  ?  : k;
         k = k > n ? n : k;
         ptr = head;
         for (i = ; i < k; ++i) {
             ptr = ptr->next;
         }
         sol.deleteThatNode(ptr);

         // print the list
         printf("%d", head->val);
         ptr = head->next;
         while (ptr != nullptr) {
             printf("->%d", ptr->val);
             ptr = ptr->next;
         }
         printf("\n");

         // delete the list
         while (head != nullptr) {
             ptr = head->next;
             delete head;
             head = ptr;
         }
     }

     return ;
 }