Ch2-2: return the nth to the last node data of a singly linked list

最新推荐文章于 2020-04-12 16:11:33 发布

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本文介绍了一种使用递归的方法来找到单链表中倒数第N个节点的数据。通过递归特性实现后进先出（LIFO）原则，文章详细展示了递归函数的设计思路及其实现代码，并对比了两种不同实现方式的效果。

In this problem, I need to find the nth to the last node data of a singly linked list.

In order to do so, one good way is to use recursion, since it is automatically put recursion part into stack and what we want is to implement LIFO.

So here it comes:

void findn(node* head){
	if(head==NULL) return;
	//head = head->next; // if use it, output n-1th not nth
	findn(head->next);
	if(nn==1) pp = head;
	--nn;
}

It is very important to fully understand recursion. A good example that I followed is from <Absolute C++, 4th>.

full code comes here, very useful to learn C++. But still lack of use of class...Need to learn it, as well as .h file.

// 2-2 tony

#include 

using namespace std;

struct node{
    int data;
    node* next;
};

node* init(int* a, int n){
	node *head, *p;
	for (int i=0; idata = a[i];
		if(i==0){
			head = p = nd;
			continue;
		}
		p->next = nd;
		p = nd;
	}
    return head;
}

void print(node* head){
    if(head==NULL) {
        cout << "end" << endl;
        return;  // must add return here so 
        			// as to jump out of recursion
    }
    cout<data<<"_ ";
    head=head->next;     
	print(head);
}

//
	int nn;
	node* pp;
//

void findn(node* head){
	if(head==NULL) return;	
	findn(head->next);
	if(nn==1) pp = head;
	--nn;
}

node* findn1(node *head, int n){
	if(head==NULL || n<1) return NULL;
	node *p, *q;
	p = q = head;
	while(n>0 && q){
		n--;
		q = q->next;
	}
	if(n>0) return NULL;
	while(q){
		p=p->next;
		q=q->next;
	}
	return p;
}

int main(){
	int n = 10;
	int a[] = {
        10,9,8,7,6,5,4,3,2,1
    };

    node* head = init(a,n);
    
    nn = 6;
    print(head);
    node* p = findn1(head, 6);
    if(p) cout<< "solution1: " <data<data << endl;
    return 0;

}

/*
node* init(int* a, int n){
	node *head, *p = head->next;
	if(n==0) return;
	else if(n==1) return;
	else{
		head->data = a[0];
		for(int i=1; i< n; ++i){
			p->data = a[i]
		}
		return head;
	}
}
*/

Executing the program....
$demo 
10_ 9_ 8_ 7_ 6_ 5_ 4_ 3_ 2_ 1_ end
solution1: 6
solution2: 6

I can also have depth variable feed as input to findn() function but should use (depth==2) so as to get the 6th to the last as output instead of 5th. need more deep understanding of recursion: chaining;;;;

void findn(node* head, int depth){
	if(head==NULL) return;
	findn(head->next, depth-1);
	if(depth==1) pp = head;  //But the index is wrong
	//--depth;
}

Here is the not correct output: should get 6 instead of 5

Executing the program....
$demo 
10_ 9_ 8_ 7_ 6_ 5_ 4_ 3_ 2_ 1_ end
5