华为上机题汇总（十一）

注：编译环境为Visual Studio 2012，答案仅供参考。

目录

第五十一题

51.子串分离
题目描述：
通过键盘输入任意一个字符串序列，字符串可能包含多个子串，子串以空格分隔。请编写一
个程序，自动分离出各个子串，并使用’,’将其分隔，并且在最后也补充一个’,’并将子
串存储。
如果输入“abc def gh i d”，结果将是abc,def,gh,i,d,
要求实现函数：
void DivideString(const char *pInputStr, long lInputLen, char *pOutputStr);
【输入】 pInputStr： 输入字符串
lInputLen： 输入字符串长度
【输出】 pOutputStr： 输出字符串，空间已经开辟好，与输入字符串等长；

#include <iostream>
using namespace std;

void DivideString(const char *pInputStr, long lInputLen, char *pOutputStr){
    while (*pInputStr != ‘\0‘)
    {
        if (*pInputStr == ‘ ‘)
        {
            pInputStr++;
            continue;
        }
        while (*pInputStr != ‘\0‘ && *pInputStr != ‘ ‘)
        {
            *pOutputStr++ = *pInputStr++;
        }
        *pOutputStr++ = ‘,‘;
    }
    *pOutputStr = ‘\0‘;
}

int main()
{
    char input[100],output[100];
    cin.getline(input,100);
    DivideString(input,strlen(input),output);
    cout << output << endl;
    return 0;
}

第五十二题

52.链表翻转。
给出一个链表和一个数k，比如链表1→2→3→4→5→6，k=2，则翻转后2→1→4→3→6→5，若k=3,翻转后3→2→1→6→5→4，若k=4，翻转后4→3→2→1→5→6，用程序实现
思想：采用遍历链表，分成length/k组，对每组进行逆转，逆转的同时要将逆转后的尾和头连接起来

#include <iostream>
using namespace std;

bool invalidInput = false;

struct LNode{
    int value;
    LNode* next;
};

LNode* reverKNodes(LNode *begin, int k){
    LNode *head = begin->next, *front = head, *p = front->next;
    for (int i = 0; i < k-1; i++)
    {
        LNode *after = p->next;
        p->next = front;
        front = p;
        p = after;
    }
    begin->next = front;
    head->next = p;
    return head;
}

LNode* reverseList(LNode *phead, int k){
    if (!phead || k < 0)
    {
        invalidInput = true;
        return phead;
    }

    LNode *p = phead;
    bool end = false;
    while (p->next != NULL)
    {
        LNode *pAhead = p;
        for (int i = 0; i < k; i++)
        {
            if (pAhead->next == NULL){
                end = true;
                break;
            }
            pAhead = pAhead->next;
        }
        if (end) break;
        p = reverKNodes(p,k);
    }
    return phead;
}

int main()
{
    LNode node6;
    node6.value = 6;
    node6.next = NULL;
    LNode node5;
    node5.value = 5;
    node5.next = &node6;
    LNode node4;
    node4.value = 4;
    node4.next = &node5;
    LNode node3;
    node3.value = 3;
    node3.next = &node4;
    LNode node2;
    node2.value = 2;
    node2.next = &node3;
    LNode node1;
    node1.value = 1;
    node1.next = &node2;
    LNode headNode;
    LNode *phead = &headNode;
    phead->next = &node1;

    int k;
    cin >> k;
    reverseList(phead,k);
    while (phead->next != NULL)
    {
        phead = phead->next;
        cout << phead->value << " ";
    }
    cout << endl;
    return 0;
}

第五十三题

53.链表相邻元素翻转
如a->b->c->d->e->f->g，翻转后变为：b->a->d->c->f->e->g

#include <iostream>
using namespace std;

bool invalidInput = false;

struct LNode{
    char value;
    LNode* next;
};

LNode* reverKNodes(LNode *begin){
    LNode *head = begin->next, *front = head, *current = front->next, *after = current->next;
    current->next = front;
    begin->next = current;
    head->next = after;
    return head;
}

LNode* reverseList(LNode *phead){
    if (!phead)
    {
        invalidInput = true;
        return phead;
    }

    LNode *p = phead;
    bool end = false;
    while (p->next != NULL)
    {
        LNode *pAhead = p;
        for (int i = 0; i < 2; i++)
        {
            if (pAhead->next == NULL){
                end = true;
                break;
            }
            pAhead = pAhead->next;
        }
        if (end) break;
        p = reverKNodes(p);
    }
    return phead;
}

int main()
{
    LNode node7;
    node7.value = ‘g‘;
    node7.next = NULL;
    LNode node6;
    node6.value = ‘f‘;
    node6.next = &node7;
    LNode node5;
    node5.value = ‘e‘;
    node5.next = &node6;
    LNode node4;
    node4.value = ‘d‘;
    node4.next = &node5;
    LNode node3;
    node3.value = ‘c‘;
    node3.next = &node4;
    LNode node2;
    node2.value = ‘b‘;
    node2.next = &node3;
    LNode node1;
    node1.value = ‘a‘;
    node1.next = &node2;
    LNode headNode;
    LNode *phead = &headNode;
    phead->next = &node1;

    reverseList(phead);
    while (phead->next != NULL)
    {
        phead = phead->next;
        cout << phead->value << " ";
    }
    cout << endl;
    return 0;
}

第五十四题

54.求最长连续子串

#include <iostream>
#include <string>
using namespace std;

void maxLengthSub(const string &s1, string &s2){
    auto begin = s1.begin();
    string maxStr;
    while (begin != s1.end())
    {
        if (*begin == ‘ ‘)
        {
            begin++;
            continue;
        }
        auto ahead = begin + 1;
        while (ahead != s1.end() && *ahead != ‘ ‘)
        {
            ahead++;
        }
        if (ahead - begin > maxStr.size())
        {
            maxStr = string(begin,ahead);
        }
        begin = ahead;
    }
    s2 = maxStr;
}

int main()
{
    string s1, s2;
    getline(cin,s1);
    maxLengthSub(s1,s2);
    cout << s2 << endl;
    return 0;
}

第五十五题

55.描述: 自从有了智能手机，时刻都要关心手机的电量。你的任务很简单，用程序打印符号来表示当前手机的电量。
用10行和10列来表示电池的电量，同时在外围加上边框，每一行表示10%的电量。
假设还有60%的电量，则显示如下：

+----------+
|----------|
|----------|
|----------|
|----------|
|++++++++++|
|++++++++++|
|++++++++++|
|++++++++++|
|++++++++++|
|++++++++++|
+----------+

运行时间限制: 无限制
内存限制: 无限制
输入: 多组测试数据，第一行为测试数据组数N（N<10），紧接着是N行，每行一个数，表示电量，这个数值可能是0，10，20 ，30，40，50，60，70，80，90，100
输出: 每组数据输出一个电池的电量，每组数据之间用15个“=”隔开
样例输入: 2
50
0

#include <iostream>
#include <string>
#include <vector>
using namespace std;

string s[] = {"+----------+","|----------|","|++++++++++|","==============="};

void display(int n){
    if (n < 0 || n > 10)
    {
        return;
    }
    cout << s[0] << endl;
    for (int i = 0; i < 10-n; i++)
    {
        cout << s[1] << endl;
    }
    for (int i = 0; i < n; i++)
    {
        cout << s[2] << endl;
    }
    cout << s[0] << endl;
}

int main()
{
    vector<int> v;
    int n;
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        int buttery;
        cin >> buttery;
        v.push_back(buttery/10);
    }
    for (unsigned i = 0; i < v.size(); i++)
    {
        display(v[i]);
        if (i != v.size()-1)
        {
            cout << s[3] << endl;
        }
    }
    cout << endl;
    return 0;
}