《算法导论》部分习题和思考题解答 - 10.2-8 单域处理前向和后向双指针

原题：说明如何对每个元素仅使用一个指针np[x](而不是两个指针next和prev)来实现双链表。假设所有指针值都是k位的整型数，且定义np[x] = next[x] XOR prev[x]，即next[x]和prev[x]的k位异或（NIL用0表示）。注意要说明访问表头所需的信息，以及如何实现在该表上的SEARCH,INSERT,和DELETE操作。如何在O(1)时间内实现这样的表。

大概思路即是x^y^x = y。因此我们只要知道前向指针的值和当前指针所存储的np[x]，即可通过异或操作求出next[x]。表头的地址存储在pHead里，因此在遍历时保存前一个节点的地址，再和当前节点的np[x]异或，就可以得到next[x]了。

代码如下：

#include <iostream>
#include <fstream>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <sstream>
#include <math.h> 
#include <string.h>
#include <algorithm>
#include <numeric>
#include <deque>
#include <climits>
#include <iomanip>
#include <unordered_map>
#include <iterator>
#include <list>
#include <chrono>
#include <unordered_set>

using namespace std;

typedef long long ll;

struct Node {
	int val;

	// '0' means no next
	int address;

	Node(int val, int address = 0) : val(val), address(address) {}
	void setAddress(int address) {
		this->address = address;
	}

	int getAddress() {
		return address;
	}
};

Node* pHead = NULL;

void insertEle(int val) {
	if (pHead == NULL) {
		pHead = new Node(val);
	}
	else {
		Node* pCurrent = new Node(val);
		Node* pNext = pHead;
		pHead = pCurrent;
		pHead->setAddress((int)pNext);
		pNext->setAddress(((int)pCurrent) ^ pNext->getAddress());
	}
}

Node* searchEle(int val) {
	Node* pEle = pHead;
	int preAddress = 0;

	while (pEle != NULL) {
		if (pEle->val == val) {
			return pEle;
		}

		int address = (int)pEle;
		pEle = (Node*)(pEle->getAddress() ^ preAddress);
		preAddress = address;
	}

	return NULL;
}

void deleteEle(Node* p) {
	Node* pEle = pHead;
	Node* pPrev = NULL;

	while (pEle != NULL) {
		if (p == pEle) {
			int nextAddress = ((int)pEle->getAddress()) ^ (int)pPrev;
			int currentAddress = (int)pEle;
			int prevAddress = (int)pPrev;

			Node* pNext = (Node*)nextAddress;
			Node* pCurrent = (Node*)currentAddress;

			if (pPrev)
				pPrev->setAddress(pPrev->getAddress() ^ currentAddress ^ nextAddress);

			if (pNext)
				pNext->setAddress(pNext->getAddress() ^ currentAddress ^ prevAddress);

			if (pHead == pCurrent) {
				pHead = pNext;
			}

			delete(pCurrent);

			break;
		}
		else {
			Node* pEleCache = pEle;
			pEle = (Node*)(pEle->getAddress() ^ (int)pPrev);
			pPrev = pEleCache;
		}
	}
}

int main() {

    // test
	insertEle(1);
	insertEle(2);
	Node* p1 = searchEle(2); 
	cout << "Search 2: " << p1->val << endl;
	Node* p2 = searchEle(3);
	cout << "Search 3: " << p2 << endl;
	deleteEle(pHead);
	Node* p3 = searchEle(2);
	cout << "Search 2: " << p3 << endl;
	Node* p4 = searchEle(1);
	cout << "Search 1: " << p4->val << endl;

    return 0;
}