分析一下这俩句的作用 cur->next = cur->next->next; struct ListNode* ans = dummy->next;
时间: 2024-06-13 10:09:01 浏览: 22
这两句代码分别来自两个不同的代码段,因此需要分别解释。
第一句代码 `cur->next = cur->next->next` 出现在链表操作中,它的作用是将当前节点的下一个节点指针指向下下个节点,相当于删除了当前节点的下一个节点。这个操作可以用于删除链表中的某个节点。
第二句代码 `struct ListNode* ans = dummy->next` 出现在链表操作中,它的作用是将链表的头节点指针指向虚拟头节点的下一个节点,也就是链表的第一个实际节点。这个操作可以用于返回链表的头节点。
相关问题
#include <iostream> using namespace std; struct ListNode { int val; ListNode* next; ListNode(int x) : val(x), next(NULL) {} }; ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { ListNode* dummy = new ListNode(0); ListNode* cur = dummy; while (l1 && l2) { if (l1->val <= l2->val) { cur->next = l1; l1 = l1->next; } else { cur->next = l2; l2 = l2->next; } cur = cur->next; } if (l1) { cur->next = l1; } else { cur->next = l2; } return dummy->next; } int main() { ListNode* l1 = new ListNode(1); l1->next = new ListNode(3); l1->next->next = new ListNode(5); ListNode* l2 = new ListNode(2); l2->next = new ListNode(4); l2->next->next = new ListNode(6); ListNode* merged = mergeTwoLists(l1, l2); while (merged) { cout << merged->val << " "; merged = merged->next; } return 0; }将这段代码中的有序表改成可自己输入
好的,你可以参考以下代码,其中cin输入的数字需要自己输入,每输入一个数字按回车键确认,输入完毕后输入-1结束输入。
```cpp
#include <iostream>
using namespace std;
struct ListNode {
int val;
ListNode* next;
ListNode(int x) : val(x), next(NULL) {}
};
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* dummy = new ListNode(0);
ListNode* cur = dummy;
while (l1 && l2) {
if (l1->val <= l2->val) {
cur->next = l1;
l1 = l1->next;
} else {
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
}
if (l1) {
cur->next = l1;
} else {
cur->next = l2;
}
return dummy->next;
}
int main() {
ListNode* l1 = new ListNode(0);
ListNode* l2 = new ListNode(0);
ListNode* cur1 = l1;
ListNode* cur2 = l2;
int num;
cout << "Please enter the numbers for the first linked list: " << endl;
while (cin >> num && num != -1) {
cur1->next = new ListNode(num);
cur1 = cur1->next;
}
cout << "Please enter the numbers for the second linked list: " << endl;
while (cin >> num && num != -1) {
cur2->next = new ListNode(num);
cur2 = cur2->next;
}
ListNode* merged = mergeTwoLists(l1->next, l2->next);
while (merged) {
cout << merged->val << " ";
merged = merged->next;
}
return 0;
}
```
希望能够帮到你!
用C语言改写,将打印输出的结果中重复的数去掉#include <stdio.h>#include <stdlib.h>struct ListNode { int val; struct ListNode *next;};struct ListNode* mergeList(struct ListNode* l1, struct ListNode* l2){ struct ListNode* dummy = (struct ListNode*)malloc(sizeof(struct ListNode)); // 新建一个虚拟头节点 dummy->val = 0; dummy->next = NULL; struct ListNode* cur = dummy; while (l1 && l2) { if (l1->val <= l2->val) { cur->next = l1; l1 = l1->next; } else { cur->next = l2; l2 = l2->next; } cur = cur->next; } if (l1) cur->next = l1; // 将剩余的节点加入到新链表中 if (l2) cur->next = l2; return dummy->next;}int main() { int n, m, num; scanf("%d%d", &n, &m); struct ListNode* l1 = (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode* l2 = (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode* p1 = l1; struct ListNode* p2 = l2; for (int i = 0; i < n; i++) { scanf("%d", &num); p1->next = (struct ListNode*)malloc(sizeof(struct ListNode)); p1->next->val = num; p1->next->next = NULL; p1 = p1->next; } for (int i = 0; i < m; i++) { scanf("%d", &num); p2->next = (struct ListNode*)malloc(sizeof(struct ListNode)); p2->next->val = num; p2->next->next = NULL; p2 = p2->next; } struct ListNode* res = mergeList(l1->next, l2->next); while (res) { printf("%d ", res->val); res = res->next; } return 0;}
以下是用 C 语言改写,并去掉重复数的代码:
#include <stdio.h>
#include <stdlib.h>
struct ListNode {
int val;
struct ListNode *next;
};
struct ListNode* mergeList(struct ListNode* l1, struct ListNode* l2){
struct ListNode* dummy = (struct ListNode*)malloc(sizeof(struct ListNode)); // 新建一个虚拟头节点
dummy->val = 0;
dummy->next = NULL;
struct ListNode* cur = dummy;
while (l1 && l2) {
if (l1->val <= l2->val) {
cur->next = l1;
l1 = l1->next;
} else {
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
}
if (l1) cur->next = l1; // 将剩余的节点加入到新链表中
if (l2) cur->next = l2;
return dummy->next;
}
int main() {
int n, m, num;
scanf("%d%d", &n, &m);
struct ListNode* l1 = (struct ListNode*)malloc(sizeof(struct ListNode));
struct ListNode* l2 = (struct ListNode*)malloc(sizeof(struct ListNode));
struct ListNode* p1 = l1;
struct ListNode* p2 = l2;
for (int i = 0; i < n; i++) {
scanf("%d", &num);
p1->next = (struct ListNode*)malloc(sizeof(struct ListNode));
p1->next->val = num;
p1->next->next = NULL;
p1 = p1->next;
}
for (int i = 0; i < m; i++) {
scanf("%d", &num);
p2->next = (struct ListNode*)malloc(sizeof(struct ListNode));
p2->next->val = num;
p2->next->next = NULL;
p2 = p2->next;
}
struct ListNode* res = mergeList(l1->next, l2->next);
// 去除重复数
int hash[10001] = {0}; // 哈希表,假设节点值范围在 [0, 10000]
struct ListNode* cur = res;
struct ListNode* pre = NULL;
while (cur) {
if (hash[cur->val]) { // 如果当前节点值已经出现过,则删除当前节点
pre->next = cur->next;
free(cur);
cur = pre->next;
} else { // 如果当前节点值没有出现过,则将其加入哈希表,并继续遍历下一个节点
hash[cur->val] = 1;
pre = cur;
cur = cur->next;
}
}
// 输出结果
cur = res;
while (cur) {
printf("%d ", cur->val);
cur = cur->next;
}
// 释放内存
cur = res;
while (cur) {
struct ListNode* temp = cur;
cur = cur->next;
free(temp);
}
return 0;
}