一.问题描述
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->NULL
, m = 2 and n = 4,
return 1->4->3->2->5->NULL
.
二.我的解题思路
现在看来,链表的题目往往是最好做的。不需要多么复杂的算法思想,只需要逻辑够严密,能够处理好各种指针边界条件即可。对于本题,维护pre,st,end,after四个指针就可以了。测试通过的程序如下:
/**
* 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) {
if(head==NULL) return NULL;
ListNode* st; ListNode* end;
int i=1;
ListNode* curr=head;ListNode* pre=NULL;
while(curr){
if(i==m) break;
pre=curr;
curr=curr->next;
i++;
}
st=curr;end=curr;ListNode* after=curr->next;
if(pre!=NULL){
for(int j=0;j<n-m;j++){
pre->next=after;
end->next=after->next;
after->next=st;
st=pre->next;
}
}
else{
for(int j=0;j<n-m;j++){
/// pre->next=after;
end->next=after->next;
after->next=st;
st=after;
}
head=after;
}
return head;
}
};