Bootstrap

C++数据结构与算法

C++数据结构与算法

1.顺序表代码模版

C++顺序表模版

#include <iostream>
using namespace std;
// 可以根据需要灵活变更类型
#define EleType int

struct SeqList
{
	EleType* elements;
	int size;
	int capacity;
};

// Init a SeqList
void InitList(SeqList* list, int capacity)
{
	list->elements = new EleType[capacity]();
	list->size = 0;
	list->capacity = capacity;
}

// Destory a SqeList
void DestoryList(SeqList* list)
{
	list->size = 0;
	delete[] list->elements;
}

// Is Empty
bool IsEmpty(SeqList* list)
{
	return list->size == 0;
}

// Inser a value into SeqList at position
void Insert(SeqList* list, int index, EleType element)
{
	// Check conditions
	if (index < 0 || index > list->size)
	{
		throw std::invalid_argument("Invalid index when insert a value into SeqList");
	}

	// Enlarge the capacity, normally enlarge 1 times
	if (list->size == list->capacity)
	{
		int newCapacity = list->capacity * 2;
		EleType* newElements = new EleType[newCapacity]();
		for (int i = 0; i < list->size; i++)
		{
			newElements[i] = list->elements[i];
		}
		delete[] list->elements;
		list->elements = newElements;
		list->capacity = newCapacity;
	}

	// Insert the data
	for (int i = list->size; i > index; --i)
	{
		list->elements[i] = list->elements[i-1];
	}
	list->elements[index] = element;
	list->size++;
}

// Delet the value at index
void DeleteElement(SeqList* list, int index)
{
	if (index < 0 || index >= list->size)
	{
		throw std::invalid_argument("Invalid index of the elements which needed delete");
		return;
	}
	for (int i = index; i < list->size - 1; i++)
	{
		list->elements[i] = list->elements[i + 1];
	}
	list->size--;
}

// Find element in list, return the index
int FindElement(SeqList* list, EleType element)
{
	for (int i = 0; i < list->size; i++)
	{
		if (list->elements[i] == element)
		{
			return i;
		}
	}
	return -1;
}

// Get element at index in the list
EleType GetElement(SeqList* list, int index)
{
	if(index < 0 || index >= list->size)
	{
		throw std::invalid_argument("Invalid index to get the element in list");
	}
	
	return list->elements[index];
}

// Update the value at index in list
void UpdateElement(SeqList* list, int index, EleType value)
{
	if (index < 0 || index >= list->size)
	{
		throw std::invalid_argument("Invalid index to Update the element in list");
	}

	list->elements[index] = value;
}

// Show
void Show(SeqList* list)
{
	if (list != NULL)
	{
		std::cout << "list size: " << list->size << std::endl;
		std::cout << "List capacity:" << list->capacity << std::endl;
		for (int i = 0; i < list->size; i++)
		{
			std::cout << "value[" << i << "] = " << list->elements[i] << " " << std::ends;
		}
		std::cout << std::endl;
	}
	else
	{
		std::cout << "The list is null" << std::endl;
	}
}

2.杭电算法2006-求奇数的乘积

1.题目–如图

在这里插入图片描述

2.使用顺序表解题代码

// 调用使用代码模版
int numbers[10000];
int main()
{
	int n;
	// 循环用例
	while(cin >> n)
	{
		SeqList list;
		InitList(&list, 1);
		int prod = 1;
		for(int i = 0 ; i < n; ++i)
		{
			int x; 
			cin >> x;
			Insert(&list, i, x);
			EleType tmp = GetElement(&list, i);
			if(tmp % 2 == 1)
			{
				prod = prod * tmp;
			}
		}
		cout << prod <<endl;
	}
	return 0;
}

3.使用数组解题

#include <iostream>
using namespace std;
int main()
{
	int n;
	while(cin >> n)
	{
		for(int i = 0; i < n; ++i)
		{
			int x;
			cin >> x;
			numbwers[i] = x;
		}
		int prod = 1;
		for(int i = 0; i < n; ++i)
		{
			int tmp = numbwers[i];
			if(tmp % 2 == 1)
			{
				prod = prod * tmp;
			}
		}
		cout << prod << endl;
	}
	return 0;
}

输入：

3 1 2 3
4 2 3 4 5

输出

3
15

3.顺序表模版更新

#include <iostream>
using namespace std;
// 可以根据需要灵活变更类型
#define EleType int

struct SeqList
{
	EleType* elements;
	int size;
	int capacity;
};

// Init a SeqList
void InitList(SeqList* list, int capacity)
{
	list->elements = new EleType[capacity]();
	list->size = 0;
	list->capacity = capacity;
}

// Destory a SqeList
void DestoryList(SeqList* list)
{
	list->size = 0;
	delete[] list->elements;
}

// Is Empty
bool IsEmpty(SeqList* list)
{
	return list->size == 0;
}

// Inser a value into SeqList at position
void Insert(SeqList* list, int index, EleType element)
{
	// Check conditions
	if (index < 0 || index > list->size)
	{
		throw std::invalid_argument("Invalid index when insert a value into SeqList");
	}

	// Enlarge the capacity, normally enlarge 1 times
	if (list->size == list->capacity)
	{
		int newCapacity = list->capacity * 2;
		EleType* newElements = new EleType[newCapacity]();
		for (int i = 0; i < list->size; i++)
		{
			newElements[i] = list->elements[i];
		}
		delete[] list->elements;
		list->elements = newElements;
		list->capacity = newCapacity;
	}

	// Insert the data
	for (int i = list->size; i > index; --i)
	{
		list->elements[i] = list->elements[i-1];
	}
	list->elements[index] = element;
	list->size++;
}

// Delet the value at index
void DeleteElement(SeqList* list, int index)
{
	if (index < 0 || index >= list->size)
	{
		throw std::invalid_argument("Invalid index of the elements which needed delete");
		return;
	}
	for (int i = index; i < list->size - 1; i++)
	{
		list->elements[i] = list->elements[i + 1];
	}
	list->size--;
}

// Find element in list, return the index
int FindElement(SeqList* list, EleType element)
{
	for (int i = 0; i < list->size; i++)
	{
		if (list->elements[i] == element)
		{
			return i;
		}
	}
	return -1;
}

// Get element at index in the list
EleType GetElement(SeqList* list, int index)
{
	if(index < 0 || index >= list->size)
	{
		throw std::invalid_argument("Invalid index to get the element in list");
	}
	
	return list->elements[index];
}

// Update the value at index in list
void UpdateElement(SeqList* list, int index, EleType value)
{
	if (index < 0 || index >= list->size)
	{
		throw std::invalid_argument("Invalid index to Update the element in list");
	}

	list->elements[index] = value;
}

// Show
void Show(SeqList* list)
{
	if (list != NULL)
	{
		std::cout << "list size: " << list->size << std::endl;
		std::cout << "List capacity:" << list->capacity << std::endl;
		for (int i = 0; i < list->size; i++)
		{
			std::cout << "value[" << i << "] = " << list->elements[i] << " " << std::ends;
		}
		std::cout << std::endl;
	}
	else
	{
		std::cout << "The list is null" << std::endl;
	}
}

4.杭电算法2008-数值统计

1.题目–如图

在这里插入图片描述

2.使用顺序表解题

#include <iostream>
using namespace std;
// 使用模版
// EleType 为double类型

int main()
{
    int n;
    while(cin >> n && n)
    {
        // 创建初始化顺序表
        SeqListlist;
        InitList(&list, 1);
        // 输出数据
        for(int i = 0; i < n; ++i)
        {
            eleType x;
            cin >> x;
            Insert(&list, i, x);
        }
        // 判断判断数据大小
        EleType tmp;
        int negative = 0;
        int positive = 0;
        int equalZero = 0;
        for(int i = 0; i < GetSize(&list); ++i)
        {
            tmp = GetElement(&list, i);
            if(tmp > 1e-8)
            {
                ++positive;
            }else if(tmp < -1e-8)
            {
                ++negative;
            }else
            {
                ++equalZero;
            }
        }
        cout << negative  <<" "<< equalZero <<" " << positive << endl;
        
    }
	return 0;
}

5.杭电算法2014-青年歌手大奖赛_评委会打分

1.题目–如图

在这里插入图片描述

2.顺序表解题

void Solution2014()
{
	int n;
	while (cin >> n)
	{
		SeqList list;
		InitList(&list, 1);

		EleType x;
		for (int i = 0; i < n; i++)
		{
			cin >> x;
			Insert(&list, i, x);
		}

		EleType max = -1000000;
		EleType min =  1000000;
		EleType sum = 0;
		for (int i = 0; i < list.size; i++)
		{
			if (max < list.elements[i])
			{
				max = list.elements[i];
			}
			if (min > list.elements[i])
			{
				min = list.elements[i];
			}
			sum += list.elements[i];
		}
		sum -= max;
		sum -= min;
		printf("%.2f\n", sum/(n-2));
	}
}
int main()
{
	Solution2014();
	return 0;
}

3.使用数组解题

#include <iostream>
using namespace std;
int main()
{
    int n;
    while(cin >> n)
    {
        doubel numbers[n];
        // 输入数据
        for(int i = 0; i < n; i++)
        {
            double x;
            cin >> x;
          	numbers[i] = x
        }
        double eleMax = -10000000;
        double eleMin = 10000000;
        double sum = 0;
        for(int i = 0; i < n; ++i)
        {
            double ele = numbers[i];
            if(ele > eleMax)
            {
                eleMax = ele;
            }
            if(ele < eleMin)
            {
                eleMin = ele;
            }
            sum += ele;
        }
		sum -= eleMax;
		sum -= eleMin;
         sum /= (n -2);
       printf("%.2f\n", sum);
        
    }
	return 0;
}

6.LeetCode-LCP 01 猜数字

1.题目–如图

在这里插入图片描述

链接： LCP 01. 猜数字 - 力扣（LeetCode）

2.解题

class Solution {
public:
    int game(vector<int>& guess, vector<int>& answer) {
        int ret = 0;
        for(int i = 0; i < 3; ++i)
        {
            if(guess[i] == answer[i])
                ret++;
        }
        return ret;
    }
};

7.LeetCode-LCP 06 拿硬币

1.题目–如图

在这里插入图片描述

2.解题:

class Solution {
public:
    int minCount(vector<int>& coins) {
        int count = 0;
        for(int i= 0; i < coins.size(); i++)
        {
            count += coins[i] / 2;
            if(coins[i]%2 == 1)
            {
                count+=1;
            }
        }
        return count;
    }
};

3.减少内存的做法

class Solution {
public:
    int minCount(vector<int>& coins) {
        int ret = 0;
        for(int i = 0; i < coins.size(); ++i)
        {
            // 加上1 后再 整除2得到结果， cpu做加法快，分支慢
            // 代替分支+/ + %的做法
            ret += (coins[i] + 1) / 2;
        }
        return ret;
    }
};

8.LeetCode-LCP 2057 值相等的最小索引

1.题目–如图

在这里插入图片描述

1.解题：

class Solution {
    public:
    int smallestEqual(vector<int>& nums) {
        int ret = -1;
        for(int i = 0; i < nums.size(); ++i)
        {
            if(i % 10 == nums[i])
            {
            	return i;
            }
        }
        return ret; 
    }
};

9.LeetCode-LCP 485 最大连续的个数

1.题目–如题

在这里插入图片描述

2.解题

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        // 00 111 01 0 1111 0000 11111111
        int ret = 0; // 最终结果
        int pre = 0; // 到当前数字最大连续1的个数（局部）
        for(int i = 0; i < nums.size(); ++i){
            if(nums[i] == 1){
                pre += 1;
                //  局部与整体比较
                if(pre > ret){
                    ret = pre;
                }
            }else{
                pre = 0;
            }
        }
        return ret;
    }
};

3.减少内存的做法

原理： 将每一个数提取出来， 并作为if的条件判断 ， 最终的落脚点是在当前的值和上一个的值的比较。

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int l = 0, r = 0;
        int maxlen = 0;
        int curlen = 0;
        while(r < nums.size())
        {
            int in = nums[r];
            r++;
            if(in)
            {
                curlen++;
            }
            else
            {
                maxlen = max(maxlen, curlen);
                curlen = 0;
                l = r;
            }
        }
        maxlen = max(maxlen, curlen);
        return maxlen;
    }
};

10.LeetCode-LCP 2006. 差的绝对值为 K 的数对数目

1.题目–如图

在这里插入图片描述

2.解题

class Solution {
public:
    int countKDifference(vector<int>& nums, int k) {
        int ret = 0;
        for(int i = 0; i < nums.size(); i++)
        {
            for(int j = i+1; j < nums.size(); j++)
            {
                if(abs(nums[i] - nums[j]) == k)
                {
                    ret++;
                }
            }
        }
        return ret;
    }
};

11.LeetCode-LCP 1464. 数组中两元素的最大乘积

1.题目–如图

在这里插入图片描述

2.解题：

暴力破解

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int ret = 0;
        int pre = 0;
        for(int i = 0; i < nums.size(); i++)
        {
            for(int j = 0; j < nums.size(); j++)
            {
                if(i != j)
                {
                    ret = (nums[i] - 1) * (nums[j] - 1);
                    if(pre < ret)
                    {
                        pre = ret;
                    }
                }
            }
        }
        return max(ret, pre);
    }
};

减少时间复杂度的做法

找到最大值的下标和次最大值的下标即可

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int maxIndex = 0;
        for(int i = 1; i < nums.size(); i ++)
        {
            if(nums[i] > nums[maxIndex])
            {
                maxIndex = i;
            }
        }
       int subMaxIndex = -1;
       for(int i = 0; i < nums.size(); i ++)
        {
           	if(i != maxIndex)
            {
                 if(nums[i] > nums[subMaxIndex])
            	{
                	subMaxIndex = i;
            	}
            }

        }
        return (num(maxIndex)-1) * (num(subMaxIndex) - 1);
    }
};

12.LeetCode-2535. 数组元素和与数字和的绝对差

1.题目–如图

在这里插入图片描述

2.解题

思路： 元素累加到x。 每一个元素的数字和累加到y， 得到x y差值的绝对值

class Solution {
public:
    int differenceOfSum(vector<int>& nums) {
        int x = 0;
        int y = 0;
        for(int i = 0; i <  nums.size(); i++)
        {
            x += nums[i];
            while(nums[i])
            {
                y += nums[i]%10;
                nums[i] /= 10;
            }
        }
        return abs(x-y);
    }
};

悦读

道可道，非常道；名可名，非常名。 无名，天地之始，有名，万物之母。 故常无欲，以观其妙，常有欲，以观其徼。 此两者，同出而异名，同谓之玄，玄之又玄，众妙之门。

最新收录
SpringBoot前后端分离项目在服务器上的部署过程详解
C++二叉搜索树
对于进程（process），你真的了解吗？
nginx 504 Gateway Time-out错误解决办法
opencascade对坐标系进行旋转操作
中国金融科技2017专题研究报告==易观智库==金融科技行业动态系列3
Linux笔记--5 --- shell脚本
Bootstrap 栅格系统初识
敏感信息泄露总结
element-ui表单验证（验证手机号是否正确，自定义验证规则）
;