[LeetCode-92] Reverse Linked List II（反转指定区间链表）

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.

【分析】反转链表，但不是从头到尾的反转，给定了一个[m,n]范围，反转指定区间的链表。1 ≤ m ≤ n ≤
length of list.不满足条件的话，返回原链表

struct ListNode* reverseList(struct ListNode* head)
{
if((head == NULL) || (head->next==NULL)) //链表为空,或只有一个结点（无需反转），直接返回
return head;

struct ListNode *pre = NULL;//前一个
struct ListNode *cur = NULL;//当前
struct ListNode *ne = NULL;//后一个

pre = head; //将前面几个节点的地址依次保存在新定义的结构体指针
cur = head ->next;

while(cur) {
ne = cur->next; //如果当前节点不为空，则将其指针域赋给ne指针
cur->next = pre; //直接将两个指针的指向反转
pre = cur; //将当前节点赋给pre，将三个指针在链表中的位子都往后移一位
cur = ne;
}
head->next = NULL;//将原来的第一个节点的指针域赋为空，作为尾节点
head = pre; //将原来的尾节点变成新链表的第一个节点
return head;
}
struct ListNode* reverseBetween(struct ListNode* head, int m, int n)
{
/*1.Exception handling*/
if(!head)
return NULL;
if(m <=0 || n<=0 || m>=n)
return head;

struct ListNode* headTemp = head;
int LinkNodeLength = 0;
struct ListNode* mStartLinkNode = head;
struct ListNode* nEndLinkNode = NULL;
struct ListNode* mStartPrivLinkNode = NULL;
struct ListNode* nEndNextLinkNode = NULL;
struct ListNode* reverseListNode = NULL;

int i = 0;

while(headTemp) {
LinkNodeLength ++;
headTemp = headTemp->next;
}
if(m>=LinkNodeLength||n>LinkNodeLength)
return head;
/*2.Find the Start of LinkNode,and save the privLinkNode of mStartLinkNode*/
for(i = 1;i<m;i++) {
mStartPrivLinkNode = mStartLinkNode;/*Save the privLinkNode of mStartLinkNode*/
mStartLinkNode = mStartLinkNode->next;
}
nEndLinkNode = mStartLinkNode;
//printf("[%d] value:%d\n",__LINE__,mStartPrivLinkNode->val);/*OK*/
//printf("[%d] value:%d\n",__LINE__,nEndLinkNode->val);/*OK*/
for(i = m;i<n;i++) {
nEndLinkNode = nEndLinkNode->next;
}
nEndNextLinkNode = nEndLinkNode->next;/*Save the nEndNextLinkNode of nEndLinkNode*/
nEndLinkNode->next = NULL;/*For the reverseList*/
//printf("[%d] value:%d\n",__LINE__,nEndLinkNode->val);
reverseListNode = reverseList(mStartLinkNode);
//printf("[%d] value:%d\n",__LINE__,reverseListNode->val);
mStartLinkNode->next = nEndNextLinkNode;/*第n节点后面一个节点*/
if(mStartPrivLinkNode) {
mStartPrivLinkNode->next = reverseListNode;
}
else {
head = reverseListNode;
}

return head;

}

全部代码如下：

// LeetCode92-Reverse Linked List II.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>
/**
* Definition for singly-linked list.
**/
struct ListNode {
int val;
struct ListNode *next;
};

struct ListNode* CreatLink(int aData[],int len)
{
if(!aData)
return NULL;
struct ListNode *head = NULL;
struct ListNode *p = NULL;
struct ListNode *Link= NULL;
/*第一个节点*/
head = (struct ListNode *)malloc(sizeof(struct ListNode));
head->val = aData[0];
head->next = NULL;
// aData ++;
Link = head; /*保存头结点*/
int i = 1;

while(i < len) {
p = (struct ListNode *)malloc(sizeof(struct ListNode));
p->val = aData[i];
head->next = p;
head = p;
i ++;
}

head->next = NULL;
return Link;
}

void printLink(struct ListNode *head)
{
while(head) {
printf("%d",head->val);
head = head->next;
}
printf("\n");
}

struct ListNode* reverseList(struct ListNode* head)
{
if((head == NULL) || (head->next==NULL)) //链表为空,或只有一个结点（无需反转），直接返回
return head;

struct ListNode *pre = NULL;//前一个
struct ListNode *cur = NULL;//当前
struct ListNode *ne = NULL;//后一个

pre = head; //将前面几个节点的地址依次保存在新定义的结构体指针
cur = head ->next;

while(cur) {
ne = cur->next; //如果当前节点不为空，则将其指针域赋给ne指针
cur->next = pre; //直接将两个指针的指向反转
pre = cur; //将当前节点赋给pre，将三个指针在链表中的位子都往后移一位
cur = ne;
}
head->next = NULL;//将原来的第一个节点的指针域赋为空，作为尾节点
head = pre; //将原来的尾节点变成新链表的第一个节点
return head;
}
struct ListNode* reverseBetween(struct ListNode* head, int m, int n)
{
/*1.Exception handling*/
if(!head)
return NULL;
if(m <=0 || n<=0 || m>=n)
return head;

struct ListNode* headTemp = head;
int LinkNodeLength = 0;
struct ListNode* mStartLinkNode = head;
struct ListNode* nEndLinkNode = NULL;
struct ListNode* mStartPrivLinkNode = NULL;
struct ListNode* nEndNextLinkNode = NULL;
struct ListNode* reverseListNode = NULL;

int i = 0;

while(headTemp) {
LinkNodeLength ++;
headTemp = headTemp->next;
}
if(m>=LinkNodeLength||n>LinkNodeLength)
return head;
/*2.Find the Start of LinkNode,and save the privLinkNode of mStartLinkNode*/
for(i = 1;i<m;i++) {
mStartPrivLinkNode = mStartLinkNode;/*Save the privLinkNode of mStartLinkNode*/
mStartLinkNode = mStartLinkNode->next;
}
nEndLinkNode = mStartLinkNode;
//printf("[%d] value:%d\n",__LINE__,mStartPrivLinkNode->val);/*OK*/
//printf("[%d] value:%d\n",__LINE__,nEndLinkNode->val);/*OK*/
for(i = m;i<n;i++) {
nEndLinkNode = nEndLinkNode->next;
}
nEndNextLinkNode = nEndLinkNode->next;/*Save the nEndNextLinkNode of nEndLinkNode*/
nEndLinkNode->next = NULL;/*For the reverseList*/
//printf("[%d] value:%d\n",__LINE__,nEndLinkNode->val);
reverseListNode = reverseList(mStartLinkNode);
//printf("[%d] value:%d\n",__LINE__,reverseListNode->val);
mStartLinkNode->next = nEndNextLinkNode;/*第n节点后面一个节点*/
if(mStartPrivLinkNode) {
mStartPrivLinkNode->next = reverseListNode;
}
else {
head = reverseListNode;
}

return head;

}

int _tmain(int argc, _TCHAR* argv[])
{
int a[] = {1,2,3,4,5,6,7,8,9,10};
struct ListNode *head1 = NULL;
struct ListNode *head2 = NULL;
head1 = CreatLink(a,sizeof(a)/sizeof(int));
printLink(head1);
head2 = reverseBetween(head1,2,6);
printLink(head2);
getchar();
return 0;
return 0;
}