数据结构与算法题目

对应知识点 https://blog.youkuaiyun.com/qq_50882544/article/details/120765639

树

二叉树

最大（小）堆

1. 修理牧场
   

1. 学到的知识点：用priority_queue优先级队列来构造最（大）小堆。

AC代码：

#include<iostream>
#include<queue>
#include<vector>
using namespace std;
int main() {
	priority_queue<int, vector<int>, greater<int>> minHeap;//优先队列里面的最小堆
	int N;
	int weight, num1, num2;
	cin >> N;//进行新一轮的输入
	for (int i = 1; i <=N; ++i) {
			cin >> num1;//逐个输入元素
			minHeap.push(num1);//将元素压入最小堆
	}
    weight = 0;
	while (minHeap.size() > 1) {//当最小堆的元素大于1
			num1 = minHeap.top();//第一个最小的元素
			minHeap.pop();//删除这个元素
			num2 = minHeap.top();//第二个最小的元素
			minHeap.pop();
			weight += num1 + num2;//获取最小的权重加和
			minHeap.push(num1 + num2);//将最小的权重加和放回到最小堆中
		}
	cout << weight << endl;
	return 0;
}

关于STL中的priority_queue优先级队列的介绍https://blog.youkuaiyun.com/qq_50882544/article/details/120913552

二叉搜索树

1. Height of BST
   

1. 学到的知识点：根据不同的序列构造BST

#include<iostream>
#include<queue>
#define MAXN 100//定义数组最大为100
using namespace std;
struct BiNode {
    int data;
    BiNode* left, * right;
};
typedef BiNode* BiTree;
void preOrder(BiTree T) {
    if (T) {
        cout << T->data << " ";
        preOrder(T->left);
        preOrder(T->right);
    }
}
void inOrder(BiTree T) {
    if (T) {
        inOrder(T->left);
        cout << T->data << " ";
        inOrder(T->right);
    }
}
void postOrder(BiTree T) {
    if (T) {
        postOrder(T->left);
        postOrder(T->right);
        cout << T->data << " ";
    }
}
void levelOrder(BiTree T) {
    if (T) {
        queue<BiTree> p;
        p.push(T);
        while (!p.empty()) {
            BiTree temp = p.front();
            cout << p.front()->data<<" ";
            p.pop();
            if (temp->left) p.push(temp->left);
            if (temp->right) p.push(temp->right);
        }
    }
    else cout << "空树" << endl;
}
int getHeight(BiTree T) {
    if (T) {
         int hl = getHeight(T->left);
         int hr = getHeight(T->right);
        if (hl > hr) return++hl;
        else return ++hr;
    }
    else return 0;
}
//根据前序序列建BST，返回树的根
BiTree creatTree(int preorder[], int start, int end) {
    if (start > end) {//如果左边查找大于右边查找
        return NULL;
    }
    if (start == end) {//有结点，直接构造结点，并返回
        BiTree tempNode = (BiTree)malloc(sizeof(struct BiNode));
        tempNode->data = preorder[start];
        tempNode->left = tempNode->right = NULL;
        return tempNode;

    }
    BiTree T = new BiNode();//用于存储根结点
    T->left = T->right = NULL;
    T->data = preorder[start];
    int index = -1;//下标
    for (int i = start + 1; i <= end; ++i) {//start值为子树的根，从其右边一个开始查找
        if (preorder[i] > preorder[start]) {//和根植比较
            index = i;//index最终停在不满足的位置
            break;
        }
    }
    //如果index为-1，证明后续值都比root的值要小，证明root只有左子树没有右子树
    if (index == -1) {
        index = end + 1;//这种情况index比所有元素都大，最后应该停在end的后一个位置
    }
    //如果根据index值，去构建当前root结点的左右子树
    T->left = creatTree(preorder, start + 1, index - 1);//左边从当前查找的后一个结点，右边有恰好满足的结点
    T->right = creatTree(preorder, index, end);//左边从正好满足的点，右边查找到末尾
    return T;
}

int main()
{
    int postorder[MAXN], N, i;
    //preorder[MAXN]为前序序列数组，N为数组个数
    cin >> N;
    for (i = 0; i < N; i++) cin >> postorder[i];
    BiTree BT = creatTree(postorder, 0,N-1);
    cout << "前序序列：";
    preOrder(BT);
    cout << endl;
    cout << "中序序列：";
    inOrder(BT);
    cout << endl;
    cout << "后序序列：";
    postOrder(BT);
    cout << endl;
    cout << "层序序列：";
    levelOrder(BT);
    cout << endl;
    cout << "树的高度："<< getHeight(BT);;
    return 0;
}
}

---

//将preorder[MAXN]替换为postorder[MAXN],前序变后序建造BST
BiTree creatTree(int postorder[], int N) {
    int i = 0, root = 0;
    BiTree T = new BiNode();
    if (N == 0) {
            return NULL;
    }
    root = postorder[N - 1];//确定根值
    T->data = root;
    //确定
    while (i < N) {
        if (postorder[i] >= root) {
            break;
        }
        i++;
    }
    T->left = creatTree(postorder, i);
    T->right = creatTree(postorder + i, N - 1 - i);
    return T;
}

---

AC代码：

#include <stdio.h>
#include <stdlib.h>
#define MAXN 10
int Height_of_BST(int preorder[], int N);
int main()
{
    int preorder[MAXN], N, i;

    scanf_s("%d", &N);
    for (i = 0; i < N; i++) scanf_s("%d", &preorder[i]);
    printf("%d\n", Height_of_BST(preorder, N));

    return 0;
}
typedef struct BiNode{
    int data;
    struct BiNode* left, *right;
}*BiTree;
BiTree creatTree(int preorder[],int start,int end) {
    if (start > end) {
        return NULL;
    }
    if (start == end) {//有结点，直接构造结点，并返回
        BiTree tempNode = (BiTree)malloc(sizeof(struct BiNode));
        tempNode->data = preorder[start];
        tempNode->left = tempNode->right = NULL;
        return tempNode;
       
    }
    BiTree T = (BiTree)malloc(sizeof(struct BiNode));//用于存储根结点
    T->left = T->right = NULL;
    T->data = preorder[start];
    int index = -1;
    for (int i = start + 1; i <= end; ++i) {
        if (preorder[i] > preorder[start]) {
            index = i;
            break;
        }
    }
    //如果index为1，证明后续值都比root的值要小，证明root只有左子树没有右子树
    if (index == -1) {
        index = end + 1;
    }
    //如果根据index值，去构建当前root结点的左右子树
    T->left = creatTree(preorder, start + 1, index - 1);
    T->right = creatTree(preorder, index, end);
    return T;
}
int getHeight(BiTree BT) {
    int hl = 0, hr = 0, height = 0;
    if (BT==NULL) {
        return 0;
    }
    hl = getHeight(BT->left);
    hr = getHeight(BT->right);
    if (hl > hr) return ++hl;
    else return ++hr;
}
int Height_of_BST(int preorder[], int N) {
    if (N== 0) {//空树返回
        return NULL;
    }
    int start = 0;
    int end = N - 1;
    BiTree BT = creatTree(preorder, start,end);
    //如果根是0层，则减1;如果为1，则不减
    return getHeight(BT)-1;
}

算法

1. 数组元素得存储位置问题

一维数组：

二维数组：

高维数组：

按行优先：

结合上边的高维数组存储图片便于理解
例如D[2][0][0]，就是2（代表两个大方格）*3（代表三行）*4（代表4列）。
按列优先：