题目描述
输入一个单向链表和一个节点的值，从单向链表中删除等于该值的节点，删除后如果链表中无节点则返回空指针。
链表结点定义如下：
struct ListNode { int m_nKey; ListNode* m_pNext; };
详细描述：
本题为考察链表的插入和删除知识。
链表的值不能重复
构造过程，例如
1 <- 2
3 <- 2
5 <- 1
4 <- 5
7 <- 2
最后的链表的顺序为 2 7 3 1 5 4
删除 结点 2
则结果为 7 3 1 5 4
#include <iostream> using namespace std; typedef struct ListNode { int m_nKey; struct ListNode* m_pNext; }ListNode; //创建链表头结点 ListNode* creatList(int headNumber) { //ListNode* pHead = new ListNode[number]; ListNode* pHead = new ListNode; pHead->m_nKey = headNumber; pHead->m_pNext = NULL; return pHead; } //插入元素 void insertList(ListNode* head, int data, int position) { ListNode* pCurrent = head; while (pCurrent->m_nKey != position) { pCurrent = pCurrent->m_pNext; } ListNode* pTemp = new ListNode; pTemp->m_nKey = data; pTemp->m_pNext = pCurrent->m_pNext; pCurrent->m_pNext = pTemp; } //删除元素//若删除后链表为空则返回空指针 ListNode* deleteList(ListNode* head, int data) { if (head->m_nKey == data && head->m_pNext == NULL) { return NULL; } if (head->m_nKey == data) { ListNode* pHead = head->m_pNext; return pHead; } ListNode* pPre = head; ListNode* pCurrent = pPre->m_pNext; while (pPre->m_pNext != NULL) { if (pCurrent->m_nKey == data) { pPre->m_pNext = pCurrent->m_pNext; } pPre = pPre->m_pNext; pCurrent = pCurrent->m_pNext; } return head; } //打印链表 void printList(ListNode* head) { ListNode* pTemp = head; while (pTemp->m_pNext != NULL) { cout << pTemp->m_nKey << " "; pTemp = pTemp->m_pNext; } cout << pTemp->m_nKey << " "; } int main() { int num; //链表节点数 while (cin >> num) { int headNumber = 0; cin >> headNumber; ListNode* pHead = creatList(headNumber); for (int i = 0; i < num - 1; i++) { int data, position; cin >> data >> position; insertList(pHead, data, position); } //printList(pHead); //cout << endl; int deldata; cin >> deldata; ListNode* pHead2 = deleteList(pHead, deldata); printList(pHead2); cout << endl; } return 0; }
