链表 2.3

实现一个算法，删除单向链表中间的某个结点，假定你只能访问该结点。

示例

输入：单向链表a->b->c->d->e中的结点c。

结果：不返回任何数据，但该链表变为：a->b->d->e。

分析：因为无法访问待删除结点的前继结点，所以只能通过复制将后续链表整体向前移动一个位置，并删除最后一个多余的结点。显然，当待删除结点为链表的尾结点时（即没有后继），该问题无解。

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

using namespace std;

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

void removeNode( Node *node );

int main( int argc, char *argv[] ) {
    string data_file = "./data.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, k = -1;
    while( ifile >>n >>k ) {
        assert( n > 0 && k > 0 && k < n );
        Node guard, *node = &guard, *target = 0;
        for( int i = 0; i < n; ++i ) {
            node->next = new Node();
            node = node->next;
            ifile >>node->val;
            cout <<node->val <<" ";
            if( i == k-1 ) { target = node; }
        }
        cout <<endl;
        if( target ) 
            cout <<"target to remove: " <<target->val <<endl;
        else
            cout <<"target to remove: null" <<endl;
        removeNode( target );
        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 removeNode( Node *node ) {
    assert( node && node->next );
    node->val = node->next->val;
    while( node->next->next ) {
        node = node->next;
        node->val = node->next->val;
    }
    delete node->next;
    node->next = 0;
    return;
}

测试文件

2 1
1 3

3 1
1 2 3

3 2
1 2 3

7 1
14 54 96 23 45 12 45

7 2
14 54 96 23 45 12 45

7 6
14 54 96 23 45 12 45