This article introduces the extended application of the classic string matching algorithm KMP
Add the fewest characters after the original string to generate a long string so that the long string contains two original strings
Time complexity: O(N)
/*
@Author: lwl2020
@Date: 2020-5-29
@Description: 在原始串的后面添加最少的字符生成一个长字符串，使长字符串包含两个原始串
*/
#include <iostream>
#include <string>
#include <vector>
using namespace std;
/*
思路：
获取字符串的next数组，这里的next数组多计算一个整个串的最长前缀和最长后缀，
将整个串的最长前缀后面的部分添加到原始串后面，即可得到包含两倍原始串的字符串
*/
class CAddChToTwice {
public:
void addChToTwice(string& str) {
if (str.empty()) {
return;
}
vector<int> next(str.length() + 1, -1);
getNextArr(str, next);
string subStr = str.substr(next.at(str.length()), str.length() - next.at(str.length()));
str.append(subStr);
}
private:
void getNextArr(string& str, vector<int>& next) {
if (str.length() == 1) {
return;
}
next.at(1) = 0;
int index = 2;
int cn = 0;
while (index <= str.length()) {
if (str.at(index - 1) == str.at(cn)) {
next.at(index++) = ++cn;
}
else if (cn == 0) {
next.at(index++) = 0;
}
else {
cn = next.at(cn);
}
}
}
};
int main() {
string str("aba");
cout << str << endl;
CAddChToTwice().addChToTwice(str);
cout << str << endl;
system("pause");
return 0;
}
Two binary trees T1 and T2, determine whether the subtree of T1 contains the same structure as T2
Time complexity: O(N)
/*
@Author: lwl2020
@Date: 2020-5-30
@Description: 两棵二叉树T1、T2，判断T1的子树是否包含与T2相同的结构
*/
#include <iostream>
#include <string>
#include <vector>
using namespace std;
struct Node {
Node* left;
Node* right;
int val;
Node(int value) :
val(value), left(nullptr), right(nullptr)
{
}
};
class CKMP {
public:
int kmp(string& str1, string& str2) {
if (str1.length() < str2.length() || str1.length() == 0 || str2.length() == 0) {
return -1;
}
vector<int> next(str2.length(), -1);
getNextArr(str2, next);
int i1 = 0;
int i2 = 0;
while (i1 < str1.length() && i2 < str2.length()) {
if (str1.at(i1) == str2.at(i2)) {
i1++;
i2++;
}
else if (i2 == 0) {
i1++;
}
else {
i2 = next.at(i2);
}
}
return i2 == str2.length() ? i1 - i2 : -1;
}
private:
void getNextArr(string& str2, vector<int>& next) {
if (str2.length() == 1) {
return;
}
next.at(1) = 0;
int index = 2;
int cn = 0;
while (index < str2.length()) {
if (str2.at(index - 1) == str2.at(cn)) {
next.at(index++) = ++cn;
}
else if (cn == 0) {
next.at(index++) = 0;
}
else {
cn = next.at(cn);
}
}
}
};
/*
思路：
先将两棵树序列化
然后用KMP算法判断是否包含
*/
class CT1ContainT2 {
public:
bool isT1ContainT2() {
Node* t1 = generateTree1();
string str1;
serialize(t1, str1);
deleteTree(t1);
Node* t2 = generateTree2();
string str2;
serialize(t2, str2);
deleteTree(t2);
return CKMP().kmp(str1, str2) != -1 ? true : false;
}
private:
void serialize(Node* t, string& str) {
if (t == nullptr) {
str.append("#");
return;
}
str.append(to_string(t->val) + "#");
serialize(t->left, str);
serialize(t->right, str);
}
Node* generateTree1() {
Node* root = new Node(1);
root->left = new Node(1);
root->left->left = new Node(1);
root->left->right = new Node(1);
root->right = new Node(1);
root->right->right = new Node(1);
return root;
}
Node* generateTree2() {
Node* root = new Node(1);
root->right = new Node(1);
return root;
}
void deleteTree(Node* root) {
if (root == nullptr) {
return;
}
deleteTree(root->left);
deleteTree(root->right);
delete root;
}
};
int main() {
cout << (CT1ContainT2().isT1ContainT2() ? "contain" : "no contain") << endl;
system("pause");
return 0;
}
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