LeetCode: Reorder List [143]

標籤：leetcode   演算法   面試   

【題目】

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-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}.

【題意】

    給定一個鏈表L: L0→L1→…→Ln-1→Ln,對他重新排序成L0→Ln→L1→Ln-1→L2→Ln-2→…

【思路】

        將鏈表對半截斷，將後半部分鏈表倒置，然後把前後兩個鏈表錯位合并
        本題涉及三個典型的鏈表操作：
        1. 找鏈表中點
        2. 鏈表倒置
        3. 鏈表歸併

【代碼】

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* reserve(ListNode *head){        if(head==NULL || head->next==NULL)return head;        ListNode* prev=NULL;        ListNode* cur=head;        ListNode* next=NULL;        while(cur){            next=cur->next;            cur->next=prev;            prev=cur;            cur=next;        }        return prev;    }    ListNode* findMid(ListNode *head){        if(head==NULL || head->next==NULL)return head;        ListNode*prev=NULL;        ListNode*p1=head;        ListNode*p2=head;        while(p2){            prev=p1;            p1=p1->next;            p2=p2->next;            if(p2)p2=p2->next;        }        prev->next=NULL;        return p1;    }    ListNode* merge(ListNode*head1, ListNode*head2){        ListNode*p1=head1;        ListNode*p2=head2;        ListNode*head=NULL, *p=NULL;        while(p2){            if(head==NULL)head=p1;             else p->next=p1;            p=p1;            p1=p1->next;            p->next=p2;            p=p2;            p2=p2->next;        }        p->next=p1;        return head;    }    void reorderList(ListNode *head) {        if(head==NULL || head->next==NULL)return;        ListNode* head2 = findMid(head);        head2 = reserve(head2);        head=merge(head, head2);    }};