int count = data.size() ;
int tag = false ; // 设置是否需要继续冒泡的标志位 for ( int i = 0 ; i < count ; i++) {
for ( int j = 0 ; j < count - i - 1 ; j++) {
if ( data[j] > data[j+1]) {
tag = true ; int temp = data[j] ; data[j] = data[j+1] ; data[j+1] = temp ; } } if ( !tag ) break ; } }
void main( void ) {
vectordata;
ifstream in(\
if ( !in) {
cout<<\ exit(1); } int temp; while (!in.eof()) {
in>>temp;
data.push_back(temp); }
in.close(); //关闭输入文件流 Order(data);
ofstream out(\ if ( !out) {
cout<<\ exit(1); }
for ( i = 0 ; i < data.size() ; i++) out< 10. 链表题：一个链表的结点结构 struct Node { int data ; Node *next ; }; typedef struct Node Node ; (1)已知链表的头结点head,写一个函数把这个链表逆序 ( Intel) Node * ReverseList(Node *head) //链表逆序 { if ( head == NULL || head->next == NULL ) return head; Node *p1 = head ; Node *p2 = p1->next ; Node *p3 = p2->next ; p1->next = NULL ; while ( p3 != NULL ) { p2->next = p1 ; p1 = p2 ; p2 = p3 ; p3 = p3->next ; } p2->next = p1 ; head = p2 ; return head ; } (2)已知两个链表head1 和head2 各自有序，请把它们合并成一个链表依然有序。(保留所有结点，即便大小相同） Node * Merge(Node *head1 , Node *head2) { if ( head1 == NULL) return head2 ; if ( head2 == NULL) return head1 ; Node *head = NULL ; Node *p1 = NULL; Node *p2 = NULL; if ( head1->data < head2->data ) { head = head1 ; p1 = head1->next; p2 = head2 ; } else { head = head2 ; p2 = head2->next ; p1 = head1 ; } Node *pcurrent = head ; while ( p1 != NULL && p2 != NULL) { if ( p1->data <= p2->data ) { pcurrent->next = p1 ; pcurrent = p1 ; p1 = p1->next ; } else { pcurrent->next = p2 ; pcurrent = p2 ; p2 = p2->next ; } } if ( p1 != NULL ) pcurrent->next = p1 ;