24.兩兩交換連結串列中的節點
題目連結：https://leetcode.cn/problems/swap-nodes-in-pairs/description/
我的程式碼：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
ListNode* dummy_head = new ListNode;
dummy_head->next = head;
ListNode* p = dummy_head;
while (p->next != nullptr && p->next->next != nullptr) {
ListNode* temp1 = p->next->next->next;
ListNode* temp2 = p->next;
p->next = p->next->next;
p->next->next = temp2;
p->next->next->next = temp1;
p = p->next->next;
}
head = dummy_head->next;
delete dummy_head;
return head;
}
};
還是定義虛擬頭節點，同時注意迴圈條件，儲存臨時節點等細節問題。
19.刪除連結串列的倒數第N個節點
題目連結：https://leetcode.cn/problems/remove-nth-node-from-end-of-list/description/
快慢指標法：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* dummy_head = new ListNode;
dummy_head->next = head;
ListNode* fast = dummy_head;
ListNode* slow = dummy_head;
while (n--) {
fast = fast->next;
}
while (fast->next) {
fast = fast->next;
slow = slow->next;
}
ListNode* p = slow->next;
slow->next = p->next;
delete p;
head = dummy_head->next;
delete dummy_head;
return head;
}
};
定義一個虛擬頭節點，先讓快指標走n步，之後兩指標同時走直到fast的next節點為空，此時慢指標指向倒數第n個節點的前一個，即可執行刪除。
面試題 02.07.連結串列相交
題目連結：https://leetcode.cn/problems/intersection-of-two-linked-lists-lcci/description/
我的程式碼：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* getIntersectionNode(ListNode* headA, ListNode* headB) {
ListNode* curA = headA;
ListNode* curB = headB;
int lengthA = 0;
int lengthB = 0;
while (curA != nullptr) {
lengthA++;
curA = curA->next;
}
while (curB != nullptr) {
lengthB++;
curB = curB->next;
}
curA = headA;
curB = headB;
if (lengthA < lengthB) {
swap(lengthA, lengthB);
swap(curA, curB);
}
int gap = lengthA - lengthB;
while (gap--) {
curA = curA->next;
}
while (curA != nullptr) {
if (curA == curB) {//注意是節點相等不是節點值相等，寫成curA->val == curB->val會答案錯誤
return curA;
}
curA = curA->next;
curB = curB->next;
}
return NULL;//不存在相應節點時別忘了返回NULL
}
};
求出兩連結串列長度，將A連結串列置為較長連結串列，然後將curA移動到與curB平齊的位置，開始比較。
142.環形連結串列II
題目連結：https://leetcode.cn/problems/linked-list-cycle-ii/description/
我的程式碼：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* detectCycle(ListNode* head) {
ListNode* fast = head;
ListNode* slow = head;
while (fast != nullptr && fast->next != nullptr) {
fast = fast->next->next;
slow = slow->next;
if (fast == slow) {
ListNode* index1 = head;
ListNode* index2 = fast;
while (index1 != index2) {
index1 = index1->next;
index2 = index2->next;
}
return index1;
}
}
return NULL;
}
};
快慢指標法判斷有無環，fast走兩步，slow走一步，若有環兩指標必相遇。
對環入口位置的數學推導：
