题目地址：https://leetcode.com/problems/reverse-linked-list-ii/description/
题目描述
Reverse a linked list from position m to n. Do it in one-pass.
Note: 1 ≤ m ≤ n ≤ length of list.
Example:
Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL
题目大意
把单链表中第m--n个元素进行翻转。
解题方法
迭代
其实就是翻转链表的而变形题目了。进行一次遍历，把第m到n个元素进行翻转，即依次插入到第m个节点的头部。
这个题还是有意思的。建议后面再多做几遍。
Python代码如下：
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseBetween(self, head, m, n):
"""
:type head: ListNode
:type m: int
:type n: int
:rtype: ListNode
"""
count = 1
root = ListNode(0)
root.next = head
pre = root
while pre.next and count < m:
pre = pre.next
count += 1
if count < m:
return head
mNode = pre.next
curr = mNode.next
while curr and count < n:
next = curr.next
curr.next = pre.next
pre.next = curr
mNode.next = next
curr = next
count += 1
return root.next
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C++代码如下：
/**
* 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) {
int pos = 1;
ListNode* dummy = new ListNode(0);
dummy->next = head;
ListNode* pre = dummy;
ListNode* cur = head;
while (cur && pos < m) {
pre = pre->next;
cur = cur->next;
pos ++;
}
ListNode* tailNode = cur;
while (cur && pos <= n) {
ListNode* nxt = cur->next;
cur->next = pre->next;
pre->next = cur;
tailNode->next = nxt;
cur = nxt;
pos ++;
}
return dummy->next;
}
};
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递归
递归解法虽然简单，但是需要对程序有深刻的认识。所以写起来并不简单。理解下面这个解法之前，最好把206. Reverse Linked Listopen in new window的递归解法弄懂。
首先要记住这个reverseBetween()函数的意义：翻转链表中的[m,n]区间的元素，并且返回新链表的头结点。
1、 那么，如果m==n，则不用翻转，直接返回原来的头即可；
2、 如果m!=1的时候，该节点不用翻转，继续翻转后面的节点；
3、 如果m==1，该节点至n节点需要翻转，使用递归先把后面的翻转，此时head->next指向了翻转部分链表的结尾，把head插入到翻转部分链表的结尾即可；
/**
* 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 (m == n)
return head;
if (m != 1) {
head->next = reverseBetween(head->next, m - 1, n - 1);
return head;
} else {
ListNode* newHead = reverseBetween(head->next, 1, n - 1);
ListNode* reversedTail = head->next->next;
head->next->next = head;
head->next = reversedTail;
return newHead;
}
}
};
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