題目描述：
反轉從位置 m 到 n 的鏈表。請使用一趟掃描完成反轉。
說明: 1 ≤ m ≤ n ≤ 鏈表長度。
示例:
輸入: 1->2->3->4->5->NULL, m = 2, n = 4
輸出: 1->4->3->2->5->NULL
解題思路1:
代碼1：
# 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):
if m == n:
return head
dummy = ListNode(-1)
dummy.next = head
a = d = dummy
for i in range(m-1):
a = a.next
for j in range(n):
d = d.next
b = a.next
c = d.next
p = b
q = b.next
while(q!=c):
o = q.next
q.next = p
p = q
q = o
b.next = c
a.next = d
return dummy.next
c++寫法
/**
* 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;
auto dummy = new ListNode(-1);
dummy -> next = head;
auto a = dummy, d = dummy;
for(int i=0;i < m-1; i ++) a = a -> next;
for(int i=0;i < n; i ++) d = d -> next;
auto b = a -> next, c = d -> next;
for(auto p=b, q = b -> next; q != c;){
auto o = q -> next;
q -> next = p;
p = q, q = o;
}
b -> next = c;
a -> next = d;
return dummy -> next;
}
};
題目來源：