C++之最长公共子序列（21）---《那些奇怪的算法》

一个char*的数组，注意其长度需要我们自己求解，另一个string类型，长度可以直接使用length方法，因此可以很棒的实现，注意，代码中的h[i][j]代表s1[0…i],s2[0…j]的最长公共子序列，我们使用动态规划方法实现！

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

int lcs_(char* s1, char* s2){
int len_s1 = 0, len_s2 = 0;
while (s1[len_s1]){
len_s1++;
}
while (s2[len_s2]){
len_s2++;
}
int h[100][100] = { 0 };
for (int i = 0; i < len_s1; i++){
if (s1[i] == s2[0]) h[i][0] = 1;
}
for (int i = 0; i <len_s2; i++){
if (s1[0] == s2[i]) h[0][i] = 1;
}
for (int i = 1; i < len_s1; i++){
for (int j = 1; j < len_s2; j++){
if (s1[i] == s2[j]) h[i][j] = h[i - 1][j - 1] + 1;
else h[i][j] = h[i - 1][j] > h[i][j - 1] ? h[i - 1][j] : h[i][j - 1];
}
}
return h[len_s1 - 1][len_s2 - 1];
}
int lcs(string s1, string s2){
int h[100][100] = { 0 };
for (int i = 0; i < s1.length(); i++){
if (s1[i] == s2[0]) h[i][0] = 1;
}
for (int i = 0; i < s2.length(); i++){
if (s1[0] == s2[i]) h[0][i] = 1;
}
for (int i = 1 ; i < s1.length(); i++){
for (int j = 1 ; j < s2.length(); j++){
if (s1[i] == s2[j]) h[i][j] = h[i - 1][j - 1] + 1;
else h[i][j] = h[i - 1][j] > h[i][j - 1] ? h[i - 1][j] : h[i][j - 1];
}
}
return h[s1.length()-1][s2.length()-1];
}
int main(){
string s1 = " world";
string s2 = "hsdfa world";

int len = lcs(s1, s2);
cout << len << endl;
return 0;
}

运行结果：