Description
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,
return 1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
The original problem is here.
The original code is here.
My Solution
I solve this problem in C++, as below:
/*
*Reverse Linked List II
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<vector>
#include<stdlib.h>
using namespace std;
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseBetween(ListNode* head, int m, int n) {
if(head == NULL || head->next == NULL)
return head;
if(m>n)
return NULL;
if(m==n)
return head;
ListNode * newHead;
ListNode * lastNode, *currNode, *nextNode, *node;
if(n<=1)
return head;
node = head;
for(int i=1;i<n;i++){
node = node->next;
}
nextNode = node->next;
if(m==1){
newHead = reverseList(head, n-m+1);
head->next = nextNode;
return newHead;
}
else{
lastNode = head;
currNode = head->next;
for(int i=2;i<m;i++){
lastNode = lastNode->next;
currNode = currNode->next;
}
newHead = reverseList(currNode,n-m+1);
currNode->next = nextNode;
lastNode->next = newHead;
}
return head;
}
ListNode* reverseList(ListNode* head, int k){
stack<ListNode*> myStack;
ListNode* newHead, *node, *nextNode;
node = head;
myStack.push(node);
for(int i=1;i<k;i++){
node = node->next;
myStack.push(node);
}
newHead = node;
myStack.pop();
while(!myStack.empty()){
nextNode = myStack.top();
myStack.pop();
node->next = nextNode;
node = node->next;
}
return newHead;
}
};
Note
To solve the problem, first, we need to get the head and tail of list need to reverse. Then, reverse the list(from m to n). Finally, join the new list to the original one.