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

2014-03-18 02:24

题目：给定一个单链表，找出倒数第K个节点。

解法：让一个指针先走K步，然后俩指针一起走到尽头。当然也可以先走到尽头数出链表的长度，然后第二次少走K步。其实耗费的工夫是一样的，但貌似总有人觉得第一种方法很巧妙很优美。

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:
     ListNode* findKthNode(ListNode *head, int k) {
         if (head == nullptr || k < ) {
             return head;
         }
         struct ListNode *p1, *p2;

         p1 = p2 = head;
         int i;
         for (i = ; i < k; ++i) {
             p2 = p2->next;
             if (p2 == nullptr) {
                 return head;
             }
         }
         while (p2 != nullptr) {
             p1 = p1->next;
             p2 = p2->next;
         }

         return p1;
     }
 };

 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;
             }
         }

         // find the kth node from the end of the list
         scanf("%d", &k);
         ptr = sol.findKthNode(head, k);
         printf("%d\n", ptr->val);

         /*
         // 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 ;
 }