动态规划------最长公共子串（连续）

采用二维数组来求解，只返回长度

int lcs(string s1,int n,string s2, int m)
{
    int result = 0;
    vector<vector<int>> c(n+1,vector<int> (m+1,0));
    
    for (int i = 1; i < n+1; i++)
        for (int j = 1; j < m+1; j++)
        {
            if (s1[i - 1] == s2[j - 1])
            {
                c[i][j] = c[i-1][j - 1] + 1;
                result = max(result, c[i][j]);
            }
                
            else c[i][j] = 0;
        }
    return result;
}

采用两个一维数组，一个存储当前值，一个存储上一轮字符串比较的值，并输出公共最长字符串

 

int lcs(string s1,int n,string s2, int m)
{
	int result = 0;//字符串中连续相等的字符数的最大值
	int pos = 0;//result出现的在字符串中的位置
	vector<int> tmp(m, 0); //保存矩阵的上一行
	vector<int> c(tmp);//保存当前行
	
	for (int i = 0; i < n; i++)
	{
		string s = s1.substr(i, 1);
		c.assign(m, 0);
		for (int j = 0; j < m; j++)
		{
			if (s2.compare(j,1,s)==0)
			{
				c[j] = (j == 0 ? 1 : (tmp[j - 1] + 1));
				if (c[j]>result)
				{
					result = c[j];
					pos = j;
				}
			}
		}
		tmp.assign(c.begin(), c.end());
	}
	for (int i = pos - result + 1; i <= pos; ++i)
		cout << s2[i] << " " ;
	cout << endl;
	return result;
}