描述
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->nullptr, m = 2 and n = 4,
return 1->4->3->2->5->nullptr.
Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list.
这是第一次实现的代码（很挫—_—）
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typedef struct ListNode
{
int _var;
struct ListNode *_next;
ListNode(int var)
:_var(var)
, _next(NULL)
{}
}node,*node_p;
class Solution
{
public:
node_p ReserveList(node_p &head,int m,int n)
{
//检查边界条件
if (head == NULL){
printf("List is NULL\n");
return NULL;
}
if (m<1||n<m){//未检查n的边界
printf("rangle is error\n");
return NULL;
}
if (n == m)
return head;
//******************
node_p prev = head;
node_p a = head;
node_p b = head;
for (int i = 2; i < m; ++i){
prev = prev->_next;
}
for (int i = 1; i < m; ++i){
a = a->_next;
}
for (int i = 1; i < n; ++i){
b = b->_next;
}
node_p tmp = new node(-1);
//a->_next = b->_next;
node_p last = a;
while (a != b){
if (m == 1)
prev = prev->_next;
else
prev->_next = a->_next;
a->_next = tmp->_next;
tmp->_next = a;
if (m == 1)
a = prev;
else
a = prev->_next;
}
if (m == 1){
prev = b->_next;
b->_next = tmp->_next;
tmp->_next = b;
last->_next = prev;
node_p Newhead = tmp->_next;
free(tmp);
return Newhead;
}
prev->_next = b->_next;
b->_next = tmp->_next;
tmp->_next = b;
last->_next = prev->_next;
prev->_next = tmp->_next;
free(tmp);
return head;
}
};这是重新写的代码（还是很挫,感觉整个人都不好了）
reverse_linklist.h:
#pragma once
#include <iostream>
#include <assert.h>
#include <stdlib.h>
using namespace std;
typedef struct ListNode
{
int _var;
ListNode *_next;
ListNode(int var)
:_var(var)
,_next(NULL)
{}
}node,*node_p;
class Solution
{
public:
node_p reverse_link(node_p &list,int m,int n)
{
//边界检查
if(list==NULL)
return NULL;
if(m<1||m>n){
cout<<"parameter error"<<endl;
return NULL;
}
if(m==n)
return list;
node dummy(-1);
node_p head=&dummy;
head->_next=list;
for(int i=0;i<m-1;++i){
head=head->_next;
}
node_p first=list;
for(int i=1;i<m;++i)
first=first->_next;
node_p second=first;
for(int i=m;i<n;++i)
second=second->_next;
node_p tmp=first;
//核心步骤
while(tmp!=second){
tmp=first->_next;
first->_next=tmp->_next;
tmp->_next=head->_next;
head->_next=tmp;
}
if(m==1)
return head->_next;
return list;
}
};test.cpp
#include "reverse_linklist.h"
using namespace std;
int main()
{
node_p n1 = new node(1);
node_p n2 = new node(2);
node_p n3 = new node(3);
node_p n4 = new node(4);
node_p n5 = new node(5);
n1->_next = n2;
n2->_next = n3;
n3->_next = n4;
n4->_next = n5;
Solution s;
node_p newhead=s.reverse_link(n1,3,5);
while (newhead != NULL){
node_p tmp = newhead;
cout<<tmp->_var<<" ";
newhead = newhead->_next;
free(tmp);
}
cout<<endl;
return 0;
}运行结果：
还是来看看人家的代码吧：
自己还是弱的很，需要更努力啦^_^
《完》
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