Reorder List

143. Reorder List

题目链接：https://leetcode.com/problems/reorder-list/#/description

题目大意：给定一个单向链表 L: L0→L1→…→Ln-1→Ln,要求返回重新组织的链表：L0→Ln→L1→Ln-1→L2→Ln-2→…。要求不能申请额外的空间，并且不能修改节点的值。例如给定{1,2,3,4}，重排后为{1,4,2,3}。

 

思路：使用快慢指针把初始链表拆分成两个长度相等的链表（如果链表节点个数为奇数，则第一个链表的节点个数多一），然后将第二个链表翻转，最后将两个链表合并成一个链表。

算法复杂度：时间复杂度为O(n),空间复杂度为O(1)

代码：

/**
 * 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) {
        auto mid = splitList(head);
        if (!mid)
            return;
        mid = reverseList(mid);
        mergeLists(head, mid);
    }
private:
    ListNode *splitList(ListNode* head) {
        if (!head)
            return head;
        ListNode *slow = head, *fast = head;
        while (fast && fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        auto mid = slow->next;
        slow->next = nullptr;
        return mid;
    }
    ListNode *reverseList(ListNode* head) {
        if (!head)
            return head;
        ListNode *pre = nullptr;
        while (head) {
            auto next = head->next;
            head->next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
    void mergeLists(ListNode *first, ListNode *second) {
        ListNode *pf = first->next, *ps = second, *pm = first;
        while (pf && ps) {
            pm->next = ps;
            ps = ps->next;
            pm->next->next = pf;
            pf = pf->next;
            pm = pm->next->next;
        }
        pm->next = pf ? pf : ps;
    }
};

评测系统上运行结果：