链表 2.1

编写代码，移除未排序链表中的重复结点。

进阶

如果不得使用临时缓冲区，该怎么解决？

分析：使用set记录已访问过的值。时间复杂度O(n*logn)，若使用unordered_set或者hash_set，则时间复杂度为O(n)。

#include <iostream>
#include <fstream>
#include <assert.h>
#include <set>

using namespace std;

struct Node {
    Node(): val(0), next(0) {}
    Node( int value ): val(value), next(0) {}
    int val;
    Node *next;
};

void removeDuplicate( Node *node );

int main( int argc, char *argv[] ) {
    string data_file = "./2.1.txt";
    ifstream ifile( data_file.c_str(), ios::in );
    if( !ifile.is_open() ) {
        fprintf( stderr, "cannot open file: %s
", data_file.c_str() );
        return -1;
    }
    int n = 0, tmp = 0;
    while( ifile >>n ) {
        assert( n >= 0 );
        Node guard, *node = &guard;
        for( int i = 0; i < n; ++i ) {
            node->next = new Node();
            node = node->next;
            ifile >>node->val;
            cout <<node->val <<" ";
        }
        cout <<endl;
        removeDuplicate( guard.next );
        node = guard.next;
        cout <<"result:" <<endl;
        while( node ) {
            Node *next = node->next;
            cout <<node->val <<" ";
            delete node;
            node = next;
        }
        cout <<endl <<endl;
    }
    ifile.close();
    return 0;
}

void removeDuplicate( Node *node ) {
    if( !node || !node->next ) { return; }
    set<int> nodeSet;
    Node *end = node, *cur = node->next;
    nodeSet.insert( node->val );
    while( cur ) {
        Node *next = cur->next;
        if( nodeSet.count( cur->val ) ) {
            delete cur;
        } else {
            nodeSet.insert( cur->val );
            end->next = cur;
            end = cur;
        }
        cur = next;
    }
    end->next = 0;
    return;
}

进阶要求不使用临时缓冲区。每次删除后续序列中与当前元素值相同的元素。时间复杂度O(n^2)

void removeDuplicate( Node *node ) {
    if( !node ) { return; }
    Node *cur = node;
    while( cur ) {
        Node *runner = cur;
        while( runner->next ) {
            if( runner->next->val == cur->val ) {
                Node *next = runner->next;
                runner->next = next->next;
                delete next;
            } else {
                runner = runner->next;
            }
        }
        cur = cur->next;
    }
    return;
}

测试文件

0

1
1

2
1 3

2
2 2

3
3 2 1

3
3 2 3

4
2 43 67 2

5
441 52 145 21 245

6
12 421 5645 33 12 421

7
14 54 96 23 45 12 45

8
12 32 34 12 12 12 12 34