最长公共子串以及子序列问题

写一个最长公共子串（子串要求连续）

动态规划法

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
int LCS1(string s1, string s2) {
	int len1 = s1.size(), len2 = s2.size();
	vector<vector<int>> ma(len1 + 1, vector<int>(len2 + 1, 0));
	int result = 0;
	for (int i = 1; i <= s1.size(); ++i) {
		for (int j = 1; j <= s2.size(); ++j) {
			if (s1[i - 1] == s2[j - 1]) {
				ma[i][j] = ma[i - 1][j - 1] + 1;
			}
			else {
				ma[i][j] = 0;   //对于求最长子串，需要连续，一旦遇到不相等的字符，就将当前位置置为0
			}
			result = max(result, ma[i][j]);
		}
		
	}
	cout << result << endl;
	return result;
}
int main() {
	string s1 = "helloWorld";
	string s2 = "loop";
	LCS(s1, s2);
	return 0;
}
//结果为2，即lo

写一个最长公共子序列（子序列不要求连续）

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
int longestCommonSubsequence(string text1, string text2) {
    if(text1.empty()||text2.empty()) return 0;
    int len1=text1.size(),len2=text2.size();
    vector<vector<int>> ma(len1+1,vector<int>(len2+1,0));
    for(int i=1;i<=len1;++i){
        for(int j=1;j<=len2;++j){
            if(text1.at(i-1)==text2.at(j-1)){
                ma[i][j]=ma[i-1][j-1]+1;
            }
            else{
                ma[i][j]=max(ma[i-1][j],ma[i][j-1]);
            }
        }
    }
    return ma[len1][len2];
}
int main() {
	string s1 = "helloWorld";
	string s2 = "loop";
	longestCommonSubsequence(s1, s2);
	return 0;
}
//结果为3，即loo