【Leetcode】92. Reverse Linked List II && 206. Reverse Linked List

  The task is reversing a list in range m to n(92) or a whole list(206).

  All in one : U need three pointers to achieve this goal. 

   1) Pointer to last value

   2) Pointer to cur p value

   3) Pointer to next value

  Here, showing my code wishes can help u.

  Of course, you also need to record four nodes in special postions. 

   1) newM  2)newN  3)beforeM  4)afterN

  These may be some complex(stupid) but it's really friend to people who are reading my code and easily understood.

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

typedef struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
}ListNode, *PNode;

void create_List(PNode head)
{
    PNode  p = head;
    int n;
    cin>>n;
    for(int i = 0; i < n ;i ++){
        int t;
        cin>>t;
        if(i == 0){
            head -> val = t;
            head -> next = NULL;
            cout<<"head is "<<head->val<<endl;
            p = head;
        }else{
            PNode newNode = (PNode) malloc(sizeof(PNode));
            newNode -> val = t;
            newNode -> next = NULL;
            p -> next = newNode;
            p = newNode;
            cout<<"p is "<<p -> val<<endl;
        }
    }
}

void display(PNode  head)
{
    PNode p = head;
    while(p){
        cout<<p->val<<" -> ";
        p = p -> next;
    }cout<<endl;
}

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if(m == n || head == NULL) return head;
        ListNode *pLast = head, *p = head -> next, *pNext = NULL;
        ListNode *newN = NULL, *newM = NULL, *beforeM = head, *afterN = NULL;
        int pos = 1;
        while(p){
            if(pos == m - 1){
                beforeM = pLast;
                pLast = p;
                p = p -> next;
            }
            else if(pos >= m && pos < n){
                pNext = p -> next;
                p -> next = pLast;
                if(pos == m){
                    pLast -> next = NULL;
                    newM = pLast;
                }
                pLast = p;
                if(pos == n - 1){
                    newN = p;
                    afterN = pNext;
                }
                p = pNext;
            }else{
                pLast = p;
                p = p -> next;
            }
            pos ++;
        }
        if( m==1 && afterN == NULL){
            head = newN;
        }else if(m == 1){
            head = newN;
            newM -> next = afterN;
        }else{
            beforeM -> next = newN;
            newM -> next = afterN;
        }
        return head;
    }

    ListNode* reverseList(ListNode* head) {
        if(head == NULL) return head;
        ListNode *pLast = head, *p = head -> next, *pNext = NULL;
        while(p){
            pNext = p -> next;
            p -> next = pLast;
            if(pLast == head){
                pLast -> next = NULL;
            }
            pLast = p;
            p = pNext;
        }
        head = pLast;
        return head;
    }
};
int main()
{
    PNode head = (PNode) malloc(sizeof(PNode));;
    create_List(head);
    cout<<"after creating , head is "<<head->val<<endl;
    display(head);
    Solution tmp = Solution();
    //tmp.reverseBetween(head, 2, 3);
    tmp.reverseList(head);
    system("pause");
    return 0;
}

 

posted on 2016-06-28 17:13  暴力的轮胎  阅读(177)  评论(0编辑  收藏  举报

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