动态规划求解最长递增子序列LIS和中序后序构建二叉树

暴露出对常见算法思维不敏锐，后续应针对该方面进行训练比如刷LeetCode。
先将两个遇到的问题LIS和中序后续构建二叉树进行分析解决。

最长递增子序列

举例：[10, 9, 2, 5, 3, 7]
对其排序得到：[2, 3, 5, 7, 9, 10]
原始序列和排序后序列求最长公共子序列LCS


代码实现：

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

int max(int a, int b)
{
	return (a > b) ? a : b;
}

// 求解m-1, n-1的LCS
int lcs(int A[], int B[], int m, int n)
{
	if (m == 0 || n == 0)
		return 0;
	if (A[m - 1] == B[n - 1])
		return lcs(A, B, m - 1, n - 1) + 1;
	else
		return max(lcs(A, B, m, n - 1), lcs(A, B, m - 1, n));
}

int main()
{
	int arrs[] = { 10, 9, 2, 5, 3, 7 };
	int copy[6];
	int len = sizeof(arrs) / sizeof(arrs[0]);
	memcpy(copy, arrs, len * sizeof(int));
	sort(copy, copy + 6);
	cout << "原始序列: ";
	for (auto tmp : arrs) {
		cout << tmp << " ";
	}
	cout << endl;
	cout << "排序序列: ";
	for (auto tmp : copy) {
		cout << tmp << " ";
	}
	cout << endl;
	cout << "The length of LCS is " << lcs(arrs, copy, 6, 6) << endl;

	return 0;
}

运行结果：

中序后序构建二叉树

提供中序和后序序列构造一棵二叉树

代码实现：

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

typedef struct BSNode {
	int val;
	BSNode* leftNode;
	BSNode* rightNode;
}BSNode;

int IndexOfMiddle(vector<int> midOrder, int value) {
	for (int index = 0; index < midOrder.size(); index++) {
		if (midOrder[index] == value) {
			return index;
		}
	}
	return -1;
}

BSNode* Solution(vector<int> midOrder, vector<int> postOrder){
	if (midOrder.size() == 0) {
		return nullptr;
	}

	// 根据后序找根节点值，并构造根节点
	int rValue = postOrder[midOrder.size() - 1];
	BSNode* root = new BSNode();
	root->val = rValue;
	cout << rValue << endl;

	// 查找根节点下标
	const int index = IndexOfMiddle(midOrder, rValue);

	// 左子树的中序和后序
	vector<int> leftMidorder;
	for (int in = 0; in < index; in++) {
		leftMidorder.push_back(midOrder[in]);
	}
	vector<int> leftPostorder;
	for (int in = 0; in < index; in++) {
		leftPostorder.push_back(postOrder[in]);
	}
	root->leftNode = Solution(leftMidorder, leftPostorder);

	// 右子树的中序和后序
	vector<int> rightMidorder;
	for (int in = index + 1; in < midOrder.size(); in++) {
		rightMidorder.push_back(midOrder[in]);
	}
	vector<int> rightPostorder;
	for (int in = index; in < postOrder.size(); in++) {
		rightPostorder.push_back(postOrder[in]);
	}
	root->rightNode = Solution(rightMidorder, rightPostorder);
	
	return root;
}

int main()
{
	vector<int> midOrder = { 3, 6 ,7, 8, 9, 11 };
	vector<int> postOrder = { 3, 7, 6, 9, 11, 8 };
	BSNode* root = Solution(midOrder, postOrder);
	return 0;
}

执行结果：

[[1]] https://blog.youkuaiyun.com/u013074465/article/details/45442067