int List: Remove( int 1)i ∥/在链表中删除第i个结点 ListNode*p s qs Int k if(1-0 ∥删除表中第1个结点 &q= first; current -first= first->link; j else i p=current; current=first; ∥/找第i-1个结点 for(k=0;k<i-1;k++) if current== NULL) break; else current= current->linkint List :: Remove ( int i ) { //在链表中删除第i 个结点 ListNode *p, *q; int k = 0; if ( i == 0 ) //删除表中第 1 个结点 { q = first; current = first = first->link; } else { p = current; current = first; //找第 i-1个结点 for ( k = 0; k < i-1; k++ ) if ( current == NULL ) break; else current = current->link;