算法题② —— 链表专栏

1. 链表数据结构

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) {}};

2. 链表的删除

2.1 移除链表元素

力扣：https://leetcode.cn/problems/remove-linked-list-elements/description/

ListNode* removeElements(ListNode* head, int val) {if(head == NULL) return NULL;ListNode *dummy= new ListNode(-1);dummy->next = head;ListNode *pre = dummy;ListNode *cur = head;while(cur){if(cur->val == val){ListNode *tmp = cur->next;pre->next = tmp;delete(cur);cur = tmp;}else{pre = cur;cur = cur->next;}}return dummy->next;
}

2.2 删除链表的倒数第N个节点

力扣：https://leetcode.cn/problems/remove-nth-node-from-end-of-list/description/

ListNode* removeNthFromEnd(ListNode* head, int n) {ListNode* dummyHead = new ListNode(-1);dummyHead->next = head;ListNode* slow = dummyHead;ListNode* fast = dummyHead;while(n-- && fast) {fast = fast->next;}fast = fast->next; // fast再提前走一步，因为需要让slow指向删除节点的上一个节点while (fast) {fast = fast->next;slow = slow->next;}slow->next = slow->next->next; ListNode *tmp = slow->next;slow->next = tmp->next;delete tmp;return dummyHead->next; 
}

3. 链表的移动

3.1 反转链表

力扣：https://leetcode.cn/problems/reverse-linked-list/description/

ListNode* reverseList(ListNode* head) {if(head == NULL) return NULL;if(head->next == NULL) return head;ListNode *pre = NULL;ListNode *cur = head;while(cur){ListNode *tmp = cur->next;cur->next = pre;pre = cur;cur = tmp;}return pre;
}

3.2 两两交换链表节点

力扣：https://leetcode.cn/problems/swap-nodes-in-pairs/description/

ListNode* swapPairs(ListNode* head) {if(head == NULL) return NULL;if(head->next == NULL) return head;ListNode *dummy = new ListNode(-1);dummy->next = head;ListNode *pre = dummy;ListNode *cur = head;while(cur && cur->next){ListNode *tmp = cur->next;cur->next = tmp->next;tmp->next = cur;pre->next = tmp;pre = cur;cur = cur->next;}return dummy->next;
}

3.3 反转链表的一部分

力扣：https://leetcode.cn/problems/reverse-linked-list-ii/description/

ListNode *reverseBetween(ListNode *head, int left, int right) {ListNode *dummy = new ListNode(-1);dummy->next = head;ListNode *pre = dummy;// 第 1 步：从虚拟头节点走 left - 1 步，来到 left 节点的前一个节点for (int i = 0; i < left - 1; i++) {pre = pre->next;}// 第 2 步：从 pre 再走 right - left + 1 步，来到 right 节点ListNode *rightNode = pre;for (int i = 0; i < right - left + 1; i++) {rightNode = rightNode->next;}// 第 3 步：切断出一个子链表（截取链表）ListNode *leftNode = pre->next;ListNode *curr = rightNode->next;pre->next = nullptr;rightNode->next = nullptr;// 第 4 步：同上一题，反转链表的子区间reverseLinkedList(leftNode);// 第 5 步：接回到原来的链表中pre->next = rightNode;leftNode->next = curr;return dummy->next;
}

3.4 K个一组反转链表

力扣：https://leetcode.cn/problems/reverse-nodes-in-k-group/description/

ListNode* reverseKGroup(ListNode* head, int k) {ListNode *dummy = new ListNode(-1);dummy->next = head;         ListNode *left = head;       //每组的第一个ListNode *right = dummy;     //每组的第k个，初始为left的前一个节点ListNode *beforePre = dummy; //反转前本组的前驱while(left){for(int i = 0; i < k; ++i){if(right->next) right = right->next;else return dummy->next;  //不足k个结点，反转结束}ListNode *beforeNext = right->next; //即为下一组的leftListNode *pre = nullptr;ListNode *cur = left;while(pre != right){ListNode *tmp = cur->next;cur->next = pre;pre = cur;cur = tmp;}beforePre->next = right;left->next = beforeNext;right = left;beforePre = left;left = beforeNext;}return dummy->next;
}

3.5 旋转链表

力扣：https://leetcode.cn/problems/rotate-list/description/

3.6 分隔链表

力扣：https://leetcode.cn/problems/partition-list/description/

4. 链表的相交

4.1 链表相交

力扣：https://leetcode.cn/problems/intersection-of-two-linked-lists-lcci/description/

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {ListNode* curA = headA;ListNode* curB = headB;int lenA = 0, lenB = 0;while (curA) { lenA++;curA = curA->next;}while (curB) { lenB++;curB = curB->next;}curA = headA;curB = headB;int gap = 0;if (lenB > lenA) {gap = lenB -lenA;while(gap--) curB = curB->next;}else{ gap = lenA - lenB;while (gap--) curA = curA->next;}while (curA) {if (curA == curB) return curA;curA = curA->next;curB = curB->next;}return NULL;
}

4.2 合并链表

力扣：https://leetcode.cn/problems/merge-two-sorted-lists/description/

4.3 环形链表

力扣：https://leetcode.cn/problems/linked-list-cycle-ii/description/

ListNode *detectCycle(ListNode *head) {ListNode* fast = head;ListNode* slow = head;while(fast && fast->next) {slow = slow->next;fast = fast->next->next;if (slow == fast) {ListNode* index1 = fast;ListNode* index2 = head;while (index1 != index2) {index1 = index1->next;index2 = index2->next;}return index2; // 返回环的入口}}return NULL;
}

判断环入口的方法：
- 相遇时，slow指针走过的节点数为: x + y， fast指针走过的节点数：x + y + n (y + z)
- 因为fast指针是一步走两个节点，slow指针一步走一个节点，所以 (x + y) * 2 = x + y + n (y + z)，得到 x = n (y + z) - y
- 整理公式之后为如下公式：x = (n - 1) (y + z) + z
- 当 n = 1 时，得到 x = z，这就意味着：从头结点出发一个指针，从相遇节点也出发一个指针，这两个指针每次只走一个节点，那么当这两个指针相遇的时候就是环形入口的节点

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