链表基础问题（C++）

一、链表反转（逆序）

设置一个new_head代表反转后的头节点

#include <stdio.h>
struct ListNode{
	int val;
	ListNode *next;
	ListNode(int x):val(x), next(NULL){}
};

class Solution{
public:
	ListNode* reverseList(ListNode* head){
		ListNode* new_head = NULL; //指向新链表头节点的指针
		while(head){
			ListNode* next = head->next; //备份head->next
			head->next = new_head;       //更新head_next
			new_head = head;             //移动new_next
			head = next;                 //遍历链表
		}
		
		return new_head;
	}
};

void test(){
	ListNode a(1);
	ListNode b(2);
	ListNode c(3);
	ListNode d(4);
	ListNode e(5);
	
	a.next = &b;
	b.next = &c;
	c.next = &d;
	d.next = &e;
	
	Solution solve;
	ListNode* head = &a;
	printf("before:\n");
	while(head){
		printf("%d->", head->val);
		head = head->next;
	}
	printf("NULL\n");
	
	head = solve.reverseList(&a);
	printf("after:\n");
	while(head){
		printf("%d->", head->val);
		head = head->next;
	}
	printf("NULL\n");
}

int main(void){
	test();
	return 0;
}

二、链表区间反转

modify_list_tail：指向当前的head，即逆置后的链表尾

pre_head：初始化开始逆置的节点前驱

#include <cstdio>
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 change_len = n - m + 1; //计算需要逆置的节点个数
		ListNode* pre_head = NULL;  //初始化开始逆置的节点的前驱
		ListNode* result = head; //最终转换后的链表头节点，非特殊情况下为head
		while(head && --m){
			pre_head = head; //记录head的前驱
			head = head->next;
		}
		
		//将modify_list_tail指向当前的head，即逆置后的链表尾
		ListNode* modify_list_tail = head;
		ListNode* new_head = NULL;
		while(head && change_len){ //逆置change_len个节点
			ListNode* next = head->next;
			head->next = new_head;
			new_head = head;
			head = next;
			change_len--; //每完成一个节点逆序，change_len--
		}
		
		modify_list_tail->next = head; //连接逆序后的链表尾和逆序部分以后的一截列表
		if(pre_head){
			pre_head->next = new_head; //将逆序链表开始的节点的前驱与逆序后的头结点链接
		}else{
			result = new_head; //如果pre_head为空，说明m==1从第一个节点开始逆序 结果即为逆序后的头节点
		}
		
		return result;
	}
};

void test(){
	ListNode a(1);
	ListNode b(2);
	ListNode c(3);
	ListNode d(4);
	ListNode e(5);
	
	a.next = &b;
	b.next = &c;
	c.next = &d;
	d.next = &e;
	
	Solution solve;
	ListNode* head = &a;
	printf("before\n");
	while(head){
		printf("%d->", head->val);
		head = head->next;
	}
	printf("NULL\n");
	
	head = solve.reverseBetween(&a, 2, 4);
	printf("after\n");
	while(head){
		printf("%d->", head->val);
		head = head->next;
	}
	printf("NULL\n");
}

int main(void){
	test();
	return 0;
}

三、链表求环

1、set的使用

法1：使用set求环起始节点（算法基础班使用的是hashSet，c++中对应unordered_set）

class Solution1{
public:
	ListNode* detectCycle(ListNode* head){
		set<ListNode*> node_set;
		while(head){
			//遍历链表，若在node_set中出现了
			if(node_set.find(head) != node_set.end()){
				return head;  //返回环的第一个节点
			}
			node_set.insert(head); //将节点插入node_set
			head = head->next;
		}
		return NULL; //没有遇到环，则返回NULL
	}
};

2、快慢指针法

快慢指针法：fast指针一次走两步，slow指针一次走一步

class Solution2{
public:
	ListNode* detectCycle2(ListNode* head){
		ListNode* fast = head;
		ListNode* slow = head;
		ListNode* meet = NULL; //相遇的节点
		while(fast){
			slow = slow->next;
			fast = fast->next;
			if(!fast){
				return NULL;  //如果fast遇到链表尾，则返回NULL
			}
			fast = fast->next; //fast再走一步
			if(fast == slow){
				meet = fast; //fast和slow相遇，记录相遇的位置
				break;
			}
		}
		if(meet == NULL){
			return NULL; //如果没有相遇则证明无环
		}
		
		while(head && meet){
			if(head == meet){  //当head和meet相遇，说明遇到环的起始位置
				return head;
			}
			head = head->next; 
			meet = meet->next;
		}
		return NULL;
	}
};

3、两种方法测试

void my_test(){
	ListNode a(1);
	ListNode b(2);
	ListNode c(3);
	ListNode d(4);
	ListNode e(5);
	
	a.next = &b;
	b.next = &c;
	c.next = &d;
	d.next = &e;
	e.next = &c;
	
	ListNode* head = &a;
	ListNode* result = NULL;
	
	printf("Solution1 after:\n");
	Solution1 solve1;
	result = solve1.detectCycle(&a);
	if(result){
		printf("环节点的值为：%d\n", result->val);
	}else{
		printf("该链表无环\n");
	}
	
	printf("--------------------------------\n");
	
	printf("Solution2 after:\n");
	Solution2 solve2;
	result = solve2.detectCycle2(&a);
	if(result){
		printf("环节点的值为：%d", result->val);
	}else{
		printf("该链表无环\n");
	}
}

四、链表partition

链表区间划分：任意给一个数字，将链表划分为三个区间（小于、等于、大于），使得链表左边的数都小于这个数，右边的数都大于这个区间

#include <cstdio>
struct ListNode{
	int val;
	ListNode* next;
	ListNode(int x):val(x), next(NULL){}
};

//定义三个区间
class Solution{
public:
	ListNode* partition(ListNode* head, int x){
		ListNode less_head(0); //三个区间链表
		ListNode equal_head(0);
		ListNode more_head(0);
		ListNode* less_ptr = &less_head;
		ListNode* equal_ptr= &equal_head;
		ListNode* more_ptr = &more_head;
		while(head){
			if(head->val < x){
				less_ptr->next = head;
				less_ptr = head;
			}else if(head->val == x){
				equal_ptr->next = head;
				equal_ptr = head;
			}else{
				more_ptr->next = head;
				more_ptr = head;
			}
			head = head->next; //遍历链表
		}
		less_ptr->next = equal_head.next;
		equal_ptr->next = more_head.next;
		more_ptr->next = NULL;
		return less_head.next;
	}
};


int main(void){
	ListNode a(1);
	ListNode b(3);
	ListNode c(5);
	ListNode d(2);
	ListNode e(6);
	ListNode f(10);
	ListNode g(5);
	ListNode h(15);
	
	a.next = &b;
	b.next = &c;
	c.next = &d;
	d.next = &e;
	e.next = &f;
	f.next = &g;
	g.next = &h;
	h.next = NULL;
	
	Solution solve;
	ListNode* head = solve.partition(&a, 5);
	
	while(head){
		printf("%d->", head->val);
		head = head->next;
	}
	printf("NULL\n");
	return 0;
}

五、复杂链表的深度拷贝

解决思路：

//STL Map的使用
#include <cstdio>
#include <map>
struct RandomListNode{
	int label;
	RandomListNode *next, *random;
	RandomListNode(int x): label(x), 
					next(NULL), random(NULL){}
};

class Solution{
public:
	RandomListNode* copyRandomList(RandomListNode* head){
		std::map<RandomListNode*, int> node_map; //地址到节点位置的map
		std::vector<RandomListNode*> node_vec;   //使用vector根据存储节点位置访问地址
		RandomListNode* ptr = head;
		int i = 0;
		while(ptr){
			node_vec.push_back(new RandomListNode(ptr->label));
			node_map[ptr] = i; //记录原始链表地址至节点位置的node_map
			ptr = ptr->next;   //遍历原始序列
			i++; //记录节点位置
		}
		node_vec.push_back(0);
		ptr = head;
		i = 0;  //再次遍历原始链表，链接新链表的next指针、random指针
		while(ptr){
			node_vec[i]->next = node_vec[i+1]; //链接新链表next指针
			if(ptr->random){ //当random指针不为空时
				int id = node_map[ptr->random]; //根据node_map确认
				node_vec[i]->random = node_vec[id]; //原链表random指针,指向的位置即为id
			}
			ptr = ptr->next;
			i++;
		}
		return node_vec[0];
	}
};

int main(void){
	RandomListNode a(1);
	RandomListNode b(2);
	RandomListNode c(3);
	RandomListNode d(4);
	RandomListNode e(5);
	a.next = &b;
	b.next = &c;
	c.next = &d;
	d.next = &e;
	
	a.random = &c;
	b.random = &d;
	c.random = &c;
	e.random = &d;
	
	Solution solve;
	RandomListNode* head = solve.copyRandomList(&a);
	while(head){
		printf("label = %d ", head->label);
		if(head->random){
			printf("rand = %d\n", head->random->label);
		}else{
			printf("rand = NULL\n");
		}
		head = head->next;
	}
	
	return 0;
}