合并两个链表
https://www.nowcoder.com/share/jump/9321389651694162925269
ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
// write code here
ListNode* dummy = new ListNode(0);
ListNode* cur = dummy;
ListNode* cur1 = pHead1;
ListNode* cur2 = pHead2;
while (cur1 != nullptr && cur2 != nullptr) {
if (cur1->val <= cur2->val) {
cur->next = cur1;
cur1 = cur1->next;
} else {
cur->next = cur2;
cur2 = cur2->next;
}
cur = cur->next;
}
if (cur1 == nullptr && cur2 != nullptr) cur->next = cur2;
else if (cur2 == nullptr && cur1 != nullptr) cur->next = cur1;
return dummy->next;
}
链表是否有环
https://www.nowcoder.com/share/jump/9321389651694163138051
class Solution {
public:
bool hasCycle(ListNode *head) {
if (head == nullptr || head->next == nullptr) return false;
ListNode* fast = head;
ListNode* slow = head;
while (fast != nullptr && fast->next != nullptr) {
fast = fast->next->next;
slow = slow->next;
if (fast == slow) return true;
}
return false;
}
};
链表中环的入口
https://www.nowcoder.com/share/jump/9321389651694163164627
第一次因为想少写几步代码，所以直接return hasCycle ? index1 : null;忘了增加判断有环再进行while循环找环入口了，所以导致报错。看来还是逻辑最重要，不能一味图方便：
ListNode* EntryNodeOfLoop(ListNode* pHead) {
if (pHead == nullptr || pHead->next == nullptr) return nullptr;
ListNode* fast = pHead;
ListNode* slow = pHead;
ListNode* index1 = pHead;
ListNode* index2;
bool hasCycle = false;
while(fast != nullptr && fast->next != nullptr) {
fast = fast->next->next;
slow = slow->next;
if (slow == fast) {
index2 = slow;
hasCycle = true;
break;
}
}
if (hasCycle) {
while (index1 != index2) {
index1 = index1->next;
index2 = index2->next;
}
return index1;
}
return nullptr;
}
删除链表的倒数第n个节点
https://www.nowcoder.com/share/jump/9321389651694164822005
思路，先反转，然后从头删除第n个，再翻转回来
ListNode* reverseList(ListNode* head) {
ListNode* prev = nullptr;
ListNode* cur = head;
while (cur != nullptr) {
ListNode* temp = cur->next;
cur->next = prev;
prev = cur;
cur = temp;
}
return prev;
}
ListNode* removeNthFromEnd(ListNode* head, int n) {
// write code here
ListNode* reverseHead = reverseList(head);
ListNode* dummy = new ListNode(0);
dummy->next = reverseHead;
ListNode* cur = reverseHead;
ListNode* pre = dummy;
while (--n) {
cur = cur->next;
pre = pre->next;
}
pre->next = cur->next;
return reverseList(dummy->next);
}
找第一个公共节点
https://www.nowcoder.com/share/jump/9321389651694165945874
ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
int len1 = 0, len2 = 0;
ListNode* cur1 = pHead1;
ListNode* cur2 = pHead2;
while (cur1 != nullptr) {
len1++;
cur1 = cur1->next;
}
while (cur2 != nullptr) {
len2++;
cur2 = cur2->next;
}
if (len1 > len2) {
swap(len1, len2);
swap(pHead1, pHead2);
}
int n = len2 - len1;
cur1 = pHead1;
cur2 = pHead2;
while (n--) {
cur2 = cur2->next;
}
while (cur1 != nullptr) {
if (cur1 == cur2) return cur1;
else {
cur1 = cur1->next;
cur2 = cur2->next;
}
}
return nullptr;
}
链表相加Ⅱ
还是喜欢用翻转来做：
ListNode* reverse(ListNode* head) {
ListNode* pre = nullptr;
ListNode* cur = head;
while (cur != nullptr) {
ListNode* temp = cur->next;
cur->next = pre;
pre = cur;
cur = temp;
}
return pre;
}
ListNode* addInList(ListNode* head1, ListNode* head2) {
head1 = reverse(head1);
head2 = reverse(head2);
ListNode* cur1 = head1;
ListNode* cur2 = head2;
ListNode* dummy = new ListNode(0);
ListNode* cur = dummy;
int index = 0, sum = 0;
while (cur1 != nullptr || cur2 != nullptr) {
int val1 = (cur1 != nullptr) ? cur1->val : 0;
int val2 = (cur2 != nullptr) ? cur2->val : 0;
sum = val1 + val2 + index;
cur->next = new ListNode(sum % 10);
index = sum / 10;
if (cur1) cur1 = cur1->next;
if (cur2) cur2 = cur2->next;
cur = cur->next;
}
if (index) cur->next = new ListNode(index);
return reverse(dummy->next);
}