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给定一个单链表 L:L0→L1→…→Ln-1→Ln ,
将其重新排列后变为: L0→Ln→L1→Ln-1→L2→Ln-2→…
你不能只是单纯的改变节点内部的值,而是需要实际的进行节点交换。
示例 1:
给定链表 1->2->3->4, 重新排列为 1->4->2->3.
示例 2:
给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.
分析:将链表的左右两边合并,合并的时候将右半部分翻转。
1 快慢指针找到切分链表
2 翻转右半部分
3 依次合并
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void reorderList(ListNode* head) {
if(head == NULL || head->next == NULL) {
return;
}
ListNode* fast = head;
ListNode* slow = head;
while(fast->next != NULL && fast->next->next != NULL) {
fast = fast->next->next;
slow = slow->next;
}
fast = slow->next;
slow->next = NULL;
ListNode dummy(0);
while(fast) {
ListNode* n = dummy.next;
dummy.next = fast;
ListNode* m = fast->next;
fast->next = n;
fast = m;
}
slow = head;
fast = dummy.next;
while(slow) {
if(fast != NULL) {
ListNode* n = slow->next;
slow->next = fast;
ListNode* m = fast->next;
fast->next = n;
fast = m;
slow = n;
} else {
break;
}
}
}
};