leetcode 92.反转链表II

leetcode 92.反转链表II

题干

给你单链表的头节点 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。
 
示例 1：
输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]

示例 2：
输入：head = [5], left = 1, right = 1
输出：[5]

提示：
链表中节点数目为 n
1 <= n <= 500
-500 <= Node.val <= 500
1 <= left <= right <= n
 
进阶： 你可以使用一趟扫描完成反转吗？

题解

先通过给定的左右界在链表中记录翻转区间的左右结点，和左结点的直接前驱和右结点的直接后继（无则置nullptr）
然后遍历翻转区间，记录当前结点的直接后继，然后将当前结点的直接后继变为直接前驱，如此不断遍历一遍即可（注意边界条件）

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int left, int right) {
        ListNode* childHead = head;
        ListNode* childTail = head;
        ListNode* childPre = nullptr;
        ListNode* childSuc = nullptr;
        while(left > 1 || right > 1){
            if(left == 2) childPre = childHead;
            if(left > 1){
                childHead = childHead->next;
                left--;
            }
            if(right > 1){
                childTail = childTail->next;
                right--;
            }
        }
        childSuc = childTail->next;
        ListNode* pre = childSuc;
        while(childHead != childSuc){
            ListNode* nextNode = childHead->next;
            childHead->next = pre;
            pre = childHead;
            childHead = nextNode;
        }
        if(childPre != nullptr) childPre->next = childTail;
        else head = childTail;
        return head;
    }
};

当然也可以严格遍历一遍，只不过记录和翻转都在一次遍历内完成了

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int left, int right) {
        ListNode* currentNode = head;
        ListNode* pre = nullptr;
        ListNode* intervalPre = nullptr;
        ListNode* intervalSuc = nullptr;
        ListNode* intervalHead = head;
        ListNode* intervalTail = head;
        bool isInInterval = false;
        int currentIndex = 1;
        while(currentNode != nullptr){
            if(currentIndex == left){
                isInInterval = true;
                intervalHead = currentNode;
            }
            if(isInInterval){
                ListNode* nextNode = currentNode->next;
                currentNode->next = pre;
                pre = currentNode;
                currentNode = nextNode;
            }else{
                intervalPre = currentNode;
                pre = currentNode;
                currentNode = currentNode->next;
            }
            currentIndex++;
            if(currentIndex == right) intervalTail = currentNode;
            if(currentIndex > right) break;
        }
        if(intervalPre == nullptr) head = intervalTail;
        else intervalPre->next = intervalTail;
        intervalHead->next = currentNode;
        return head;
    }
};