Leetcode 之 Reverse Linked List II
2013-11-18 23:58
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原题:
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given
return
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
这个题比较简单,就是把第m到n个元素倒转过来。。。。看到这个题第一反应用栈来解决。。。
而且为了方便,其实在很多链表的相关实现时,都可以声明一个ListNode*叫做pre,pre的next指向head
基本思想:
1 把current一直向后,直到current的next要入栈
2 current的next开始,入栈m-n+1个元素
3 出栈,current的next指向栈顶,最后把current指向tail,即栈里最后一个元素的next指向tail
代码(20ms):
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.
这个题比较简单,就是把第m到n个元素倒转过来。。。。看到这个题第一反应用栈来解决。。。
而且为了方便,其实在很多链表的相关实现时,都可以声明一个ListNode*叫做pre,pre的next指向head
基本思想:
1 把current一直向后,直到current的next要入栈
2 current的next开始,入栈m-n+1个元素
3 出栈,current的next指向栈顶,最后把current指向tail,即栈里最后一个元素的next指向tail
代码(20ms):
class Solution { public: ListNode *reverseBetween(ListNode *head, int m, int n) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. ListNode * pre = new ListNode(-1); pre->next = head; //pre指向head的前一个 ListNode *current = pre; //把current指向第m个元素的前一个 for(int i = 0 ;i < m-1 ;i++){ current = current->next; } ListNode * tail = current->next; stack<ListNode*>s; //把第m到n个元素都入栈 //tail最后指向第n个元素的next for(int i = 0 ; i <= n-m ;i++){ ListNode* temp = tail; s.push(temp); tail = tail->next; } //把current指向栈顶 for(int i = 0; i<= n-m;i++){ current->next = s.top(); s.pop(); current = current->next; } //current的next指向原来第n个元素的next current->next = tail; return (pre->next); } };
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