鏈表 面試題整理

轉自http://blog.csdn.net/fatshaw/article/details/6452460?reload

看完覺得獲益匪淺啊，分享過來一下~~

 

1.已知鏈表的頭結點head,寫一個函數把這個鏈表逆序

void List::reverse()
{
        list_node * p = head;
        list_node * q = p->next;
        list_node * r = NULL;
        while(q){
                r = q->next;
                q->next = p;
                p = q;
                q = r;
        }
        head->next  = NULL;
        head = p;
}

void List::reverse() { list_node * p = head; list_node * q = p->next; list_node * r = NULL; while(q){ r = q->next; q->next = p; p = q; q = r; } head->next = NULL; head = p; }

遞歸方法：

void List::reverse2(list_node * curnode)
{
        if(curnode ==NULL)curnode = head;
        if(curnode->next==NULL)
        {
                cur = curnode;
                return;
        }

        reverse2(curnode->next);
        curnode->next->next=curnode;
        if(curnode == head)
        {
                head=cur;
                cur = curnode;
                cur->next = NULL;
        }
}

void List::reverse2(list_node * curnode) { if(curnode ==NULL)curnode = head; if(curnode->next==NULL) { cur = curnode; return; } reverse2(curnode->next); curnode->next->next=curnode; if(curnode == head) { head=cur; cur = curnode; cur->next = NULL; } }

2.已知兩個鏈表head1 和head2 各自有序，請把它們合併成一個鏈表依然有序。

void List::merge(List & list)
{
        list_node * a = head;
        list_node * b = list.head;
        list_node * tempa= a;
        list_node * tempb = b;
        while(a&&b)
        {
                if(a->value <= b->value)
                {
                        while(a&&b&&a->value <= b->value)
                        {
                                tempa = a;
                                a = a->next;
                        }
                        tempa->next=b;
                }
                else
                {
                        while(a&&b&&a->value > b->value)
                        {
                                tempb = b;
                                b = b->next;
                        }
                        tempb->next=a;
                }

        }
}

void List::merge(List & list) { list_node * a = head; list_node * b = list.head; list_node * tempa= a; list_node * tempb = b; while(a&&b) { if(a->value <= b->value) { while(a&&b&&a->value <= b->value) { tempa = a; a = a->next; } tempa->next=b; } else { while(a&&b&&a->value > b->value) { tempb = b; b = b->next; } tempb->next=a; } } }

遞歸方法:

list_node* List::recursive_merge(list_node * a,list_node * b)
{
        if(a == NULL)return b;
        if(b == NULL)return a;
        if(a->value <= b->value){
                a->next=recursive_merge(a->next,b);
                return a;
        }
        if(a->value > b->value)
        {
                b->next=recursive_merge(a,b->next);
                return b;
        }
}

list_node* List::recursive_merge(list_node * a,list_node * b) { if(a == NULL)return b; if(b == NULL)return a; if(a->value <= b->value){ a->next=recursive_merge(a->next,b); return a; } if(a->value > b->value) { b->next=recursive_merge(a,b->next); return b; } }

3.有一個鏈表L,其每個節點有2個指針，一個指針next指向鏈表的下個節點，另一個random隨機指向鏈表中的任一個節點，可能是自己或者爲空，寫一個程序，要求複製這個鏈表的結構並分析其複雜性

這個題目的方法很巧妙，將兩個鏈表連接起來形成一個鏈表，設置好random指針後再將兩個鏈表分割開來。

List List::copyRndList()
{
        list_node * newhead = NULL;
        list_node * newcur = NULL;
        list_node * cur = head;
        while(cur)
        {
                if(newhead == NULL){
                        newhead = new list_node;
                        newhead->value = cur->value;
                        newhead->next = cur->next;
                        cur->next = newhead;
                }
                else{
                        newcur = new list_node;
                        newcur->value = cur->value;
                        newcur->next = cur->next;
                        cur->next = newcur;
                }
                cur=cur->next->next;
        }
        cur = head;
        while(cur){
                if(cur->rnd)
                        cur->next->rnd=cur->rnd->next;
                else
                        cur->next->rnd = NULL;
                cur = cur->next->next;
        }
        cur = head;
        list_node * dst = cur->next;
        while(cur)
        {
                list_node * temp = dst->next;
                cur->next = temp;
                if(temp)dst->next = temp->next;
                cur = cur->next;
                dst=dst->next;
        }
        List newList;
        newList.head = newhead;
        return newList;
}

List List::copyRndList() { list_node * newhead = NULL; list_node * newcur = NULL; list_node * cur = head; while(cur) { if(newhead == NULL){ newhead = new list_node; newhead->value = cur->value; newhead->next = cur->next; cur->next = newhead; } else{ newcur = new list_node; newcur->value = cur->value; newcur->next = cur->next; cur->next = newcur; } cur=cur->next->next; } cur = head; while(cur){ if(cur->rnd) cur->next->rnd=cur->rnd->next; else cur->next->rnd = NULL; cur = cur->next->next; } cur = head; list_node * dst = cur->next; while(cur) { list_node * temp = dst->next; cur->next = temp; if(temp)dst->next = temp->next; cur = cur->next; dst=dst->next; } List newList; newList.head = newhead; return newList; }

4. 找出單向鏈表中中間結點

兩個指針，一個步長爲1，另一個步長爲2.步長爲2的走到底後步長爲1的正好到中間。

list_node * List::middleElement()
{
        list_node * p = head;
        list_node * q = head->next;
        while(q){
                p = p->next;
                if(q)q=q->next;
                if(q)q=q->next;
        }
}

list_node * List::middleElement() { list_node * p = head; list_node * q = head->next; while(q){ p = p->next; if(q)q=q->next; if(q)q=q->next; } }

5. 如何檢查一個單向鏈表上是否有環

同樣兩個指針，一個步長爲1，另一個步長爲2，如果兩個指針能相遇則有環。

list_node * List::getJoinPointer()
{

        if(head == NULL || head->next == NULL)return NULL;
        list_node * one = head;
        list_node * two = head->next;
        while(one != two){
                one = one->next;
                if(two)two=two->next;
                else break;
                if(two)two=two->next;
                else break;
        }
        if(one == NULL || two == NULL)return NULL;
        return one;
}

list_node * List::getJoinPointer() { if(head == NULL || head->next == NULL)return NULL; list_node * one = head; list_node * two = head->next; while(one != two){ one = one->next; if(two)two=two->next; else break; if(two)two=two->next; else break; } if(one == NULL || two == NULL)return NULL; return one; }

6. 給定單鏈表(head)，如果有環的話請返回從頭結點進入環的第一個節點。

設鏈表頭到環入口節點距離爲x，環入口節點到兩個指針相遇節點距離爲z，換長度爲y，則有x+z+1=y,所以z=y-1-x，即一個指針從鏈表頭部開始移動，一個指針兩個指針相遇後一個節點開始移動，相遇的地方即爲環入口

list_node * List::findCycleEntry()
{
        if(checkCycle()==false)return NULL;
        list_node * joinPointer = getJoinPointer();
        list_node * p = head;
        list_node * q = joinPointer->next;
        while(p!=q)
        {
                p=p->next;
                q=q->next;
        }
        return p;
}

list_node * List::findCycleEntry() { if(checkCycle()==false)return NULL; list_node * joinPointer = getJoinPointer(); list_node * p = head; list_node * q = joinPointer->next; while(p!=q) { p=p->next; q=q->next; } return p; }

7.只給定單鏈表中某個結點p(並非最後一個結點，即p->next!=NULL)指針，刪除該結點。

將p後面那個節點的值複製到p，刪除p後面的節點

void List::deleteByPointer(list_node * node)
{
        if(node)
        {
                if(node->next){
                        node->value = node->next->value;
                        node->next = node->next->next;
                }
        }
}

void List::deleteByPointer(list_node * node) { if(node) { if(node->next){ node->value = node->next->value; node->next = node->next->next; } } }

8.在p前面插入一個節點

在p後面插入新節點，將p的值與新建的節點值互換。

9.給定單鏈表頭結點，刪除鏈表中倒數第k個結點

一個指針指向鏈表頭，另一個指針指向第k個指針，然後兩個指針一起移動，第二個指針到了末端則第一個指針就是倒數第k個節點

list_node * List::lastKelement(int k){
        int t = k;
        list_node * p = head;
        while(p&&t){
                p=p->next;
                t--;
        }
        if(p == NULL && t >0)return NULL;
        list_node * q=head;
        while(q && p){
                p=p->next;
                q=q->next;
        }
        return q;
}

list_node * List::lastKelement(int k){ int t = k; list_node * p = head; while(p&&t){ p=p->next; t--; } if(p == NULL && t >0)return NULL; list_node * q=head; while(q && p){ p=p->next; q=q->next; } return q; }

10. 判斷兩個鏈表是否相交。

兩種情況，如果鏈表有環，則先在環裏設定一個指針不動，另一個鏈表從頭開始移動，如果另一個鏈表能夠與環中的指針相遇則是相交。

如果沒有環，則判斷兩個鏈表的最後個節點是否相同，相同則相交

bool List::isIntersecting(const List & list)
{
        bool flag = 
false;
        if(this->checkCycle())
        {
                list_node * p = getJoinPointer();
                list_node * q = list.head;
                while(q){
                        if(q == p){
                                flag = true;
                                break;
                        }
                        q=q->next;
                }
                flag = true;
        }
        else
        {
                list_node * p = head;
                list_node * q = list.head;
                while(p->next)p=p->next;
                while(q->next)q=q->next;
                if(p  == q)flag =
true;
                else flag =false;
        }
        return flag;
}

bool List::isIntersecting(const List & list) { bool flag = false; if(this->checkCycle()) { list_node * p = getJoinPointer(); list_node * q = list.head; while(q){ if(q == p){ flag = true; break; } q=q->next; } flag = true; } else { list_node * p = head; list_node * q = list.head; while(p->next)p=p->next; while(q->next)q=q->next; if(p == q)flag = true; else flag =false; } return flag; }

11. 兩個鏈表相交，找出交點

求出兩個鏈表的長度a和b，一個指針指向較短鏈表的頭head，另一個指針指向較長鏈表的第head+|a-b|，然後兩個指針一起移動，相遇處即爲交點。

list_node * List::intersectNode(const List & list)
{
        if(!isIntersecting(list))return NULL;
        int a = cnt;
        int b = list.cnt;
        list_node * p;
        list_node * q;
        if(a<b){p=list.head;q = head;}
        else {p = head; q=list.head;}
        a = abs(cnt - list.cnt);
        while(p && a)
        {
                p = p->next;
                a--;
        }
        while(p&&q)
        {
                if(q==p)break;
                p=p->next;
                q=q->next;
        }
        if(p && q && p == q)return p;
        return NULL;
}