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。
1 #include <iostream> 2 #include <fstream> 3 #include <assert.h> 4 5 using namespace std; 6 7 struct Node { 8 Node(): val(0), next(0) {} 9 Node( int value ): val(value), next(0) {} 10 int val; 11 Node *next; 12 }; 13 14 void removeNode( Node *node ); 15 16 int main( int argc, char *argv[] ) { 17 string data_file = "./data.txt"; 18 ifstream ifile( data_file.c_str(), ios::in ); 19 if( !ifile.is_open() ) { 20 fprintf( stderr, "cannot open file: %s ", data_file.c_str() ); 21 return -1; 22 } 23 int n = 0, k = -1; 24 while( ifile >>n >>k ) { 25 assert( n > 0 && k > 0 && k < n ); 26 Node guard, *node = &guard, *target = 0; 27 for( int i = 0; i < n; ++i ) { 28 node->next = new Node(); 29 node = node->next; 30 ifile >>node->val; 31 cout <<node->val <<" "; 32 if( i == k-1 ) { target = node; } 33 } 34 cout <<endl; 35 if( target ) 36 cout <<"target to remove: " <<target->val <<endl; 37 else 38 cout <<"target to remove: null" <<endl; 39 removeNode( target ); 40 node = guard.next; 41 cout <<"result:" <<endl; 42 while( node ) { 43 Node *next = node->next; 44 cout <<node->val <<" "; 45 delete node; 46 node = next; 47 } 48 cout <<endl <<endl; 49 } 50 ifile.close(); 51 return 0; 52 } 53 54 void removeNode( Node *node ) { 55 assert( node && node->next ); 56 node->val = node->next->val; 57 while( node->next->next ) { 58 node = node->next; 59 node->val = node->next->val; 60 } 61 delete node->next; 62 node->next = 0; 63 return; 64 }
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