數據結構問題之重排鏈表

題目

Given a singly linked list L: L 0→L 1→…→L n-1→L n,
reorder it to: L 0→L n →L 1→L n-1→L 2→L n-2→…

You must do this in-place without altering the nodes’ values.

For example,
Given{1,2,3,4}, reorder it to{1,4,2,3}.

解析

解決這個問題首先使用快慢指針找到數列中點，然後將後半部分反序，最後將前半部分和後半部分進行連接。

c++實現

/**
 * 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 || !head->next) return;

        ListNode* fast = head, *slow = head;
        while (fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        // 將slow指針後的鏈表反序連接
        ListNode* pre = slow->next;
        if (pre->next) {
            ListNode* nex = pre->next;
            pre->next = NULL; // 將原鏈表斷開，如不斷開，將會死循環
            while (nex) {
                pre = nex->next;
                nex->next = slow->next;
                slow->next = nex;
                nex = pre;
            }
        }

        // 將head指針和slow指針連接
        ListNode* p = head, *q = slow->next;
        while (p && q) {
            slow->next = q->next;
            q->next = p->next;
            p->next = q;
            q = slow->next;
            p = p->next->next;
        }
    }
};