《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) == 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;
            }
        }
        
        // remove a node from middle of the list
        scanf("%d", &k);
        k = k < 1 ? 1 : k;
        k = k > n ? n : k;
        ptr = head;
        for (i = 1; 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 0;
}