重排链表

标签：ListNode struct next 链表 重排 curr NULL

 解题思路：采用快慢指针的方法将单链表分成两半，再对两部分的链表进行合并

void reorderList(struct ListNode* head) {     if (head == NULL || head->next == NULL) {         return;     }         // 找到链表中间节点     struct ListNode* slow = head;     struct ListNode* fast = head;         while (fast != NULL && fast->next != NULL) {         slow = slow->next;         fast = fast->next->next;     }         // 翻转后半部分链表     struct ListNode* prev = NULL;     struct ListNode* curr = slow;     struct ListNode* nextTemp;         while (curr != NULL) {         nextTemp = curr->next;         curr->next = prev;         prev = curr;         curr = nextTemp;     }         // 合并两个链表     struct ListNode* firstHalf = head;     struct ListNode* secondHalf = prev;         while (secondHalf->next != NULL) {         struct ListNode* firstNextTemp = firstHalf->next;         struct ListNode* secondNextTemp = secondHalf->next;         firstHalf->next = secondHalf;         secondHalf->next = firstNextTemp;         firstHalf = firstNextTemp;         secondHalf = secondNextTemp;     } }

标签：ListNode,struct,next,链表,重排,curr,NULL
From： https://www.cnblogs.com/ymy1/p/17694492.html