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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 intstruct 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 conditionsif (index < 0 || index > list->size){throw std::invalid_argument("Invalid index when insert a value into SeqList");}// Enlarge the capacity, normally enlarge 1 timesif (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 datafor (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 intstruct 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 conditionsif (index < 0 || index > list->size){throw std::invalid_argument("Invalid index when insert a value into SeqList");}// Enlarge the capacity, normally enlarge 1 timesif (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 datafor (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 11111111int 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);}
};
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