《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 < 1) {
            return head;
        }
        struct ListNode *p1, *p2;
        
        p1 = p2 = head;
        int i;
        for (i = 0; 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) == 1 && n > 0) {
        // create a linked list
        ptr = head = nullptr;
        for (i = 0; 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 0;
}