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