您的位置:首页 > 其它

【leetcode】【单链表】【92】Reverse Linked List II

2015-05-10 10:22 323 查看
#include<iostream>
using namespace std;

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;
		ListNode* first = head;
		ListNode* preFirst = head;
		ListNode* last = head;
		ListNode* cur = head;
		while (--m > 0){  
			preFirst = first;
			first = first->next;
		}
		while (n-- > 0)
			last = last->next;//调整区域是前闭后开区间
		if (first == preFirst){ //头结点就要调整
			cur = first->next;
			first->next = last;
			while (cur != last){
				ListNode* temp = cur;
				cur = cur->next;
				temp->next = head;
				head = temp;
			}
		}else{	//头结点不需调整
			preFirst->next = last;
			cur = first->next;
			ListNode* newHead = first;
			while (cur != last){
				ListNode* temp = cur;
				cur = cur->next;
				temp->next = newHead;
				newHead = temp;
			}
			first->next = preFirst->next;
			preFirst->next = newHead;
		}
		return head;
	}
	ListNode* createList(ListNode* head){
		int numOfNode;
		int value;
		cout << "please input number of listNode:";
		cin >> numOfNode;
		cin >> value;
		head = new ListNode(value);
		ListNode* cur = head;
		for (int i = 1; i < numOfNode; ++i){
			cin >> value;
			ListNode* temp = new ListNode(value);
			cur->next = temp;
			cur = temp;
		}
		return head;
	}
	void printNode(ListNode* head){
		ListNode* cur = head;
		while (cur){
			cout << cur->val << " ";
			cur = cur->next;
		}
		cout << endl;
	}
};

int main(){
	ListNode* head = NULL;
	Solution solution;
	head = solution.createList(head);
	solution.printNode(head);

	head = solution.reverseBetween(head, 1, 2);
	solution.printNode(head);

	system("pause");
	return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: