【LCS系列】最长公共子序列和最长公共子串

最长公共子序列：

如果要回溯出整个字符串的答案的话，可以直接看dp[i][len2-1]列，或者dp[len1-1][i]这一行，变化的时候，则代表要选这个字符，然后连起来就可以了。（即构造字符串的过程是On的） 

class Solution {
public:
    /**
     * longest common substring
     * @param str1 string字符串 the string
     * @param str2 string字符串 the string
     * @return string字符串
     */
    int dp[5005][5005];
    string LCS(string str1, string str2) {
        // write code here
        int len1 = str1.size();
        int len2 = str2.size();
        for(int i = 0; i<len1; i++) dp[i][0] = (str1[i] == str2[0])?1:dp[i-1][0];
        for(int i = 0; i<len2; i++) dp[0][i] = (str1[0] == str2[i])?1:dp[0][i-1];
        for(int i = 1; i<len1; i++) {
            for(int j = 1; j<len2; j++) {
                if(str1[i] == str2[j]) dp[i][j] = dp[i-1][j-1] + 1;
                else dp[i][j] = max(dp[i-1][j-1], max(dp[i-1][j], dp[i][j-1]));
            }
        }
        for(int i = 0; i<len1; i++) {
            for(int j = 0; j<len2; j++) {
                cout << dp[i][j] << " ";
            }
            cout << endl;
        }
        cout << dp[len1-1][len2-1] <<endl;
        return str1;
    }
};

最长公共子串：

class Solution {
public:
    /**
     * longest common substring
     * @param str1 string字符串 the string
     * @param str2 string字符串 the string
     * @return string字符串
     */
    int dp[5005][5005];
    string LCS(string str1, string str2) {
        // write code here
        int len1 = str1.size();
        int len2 = str2.size();
        for(int i = 0; i<len1; i++) dp[i][0] = (str1[i] == str2[0])?1:0;
        for(int i = 0; i<len2; i++) dp[0][i] = (str1[0] == str2[i])?1:0;
        for(int i = 1; i<len1; i++) {
            for(int j = 1; j<len2; j++) {
                if(str1[i] == str2[j]) dp[i][j] = dp[i-1][j-1] + 1;
                else dp[i][j] = 0;
            }
        }
        int mx = 0;
        string ans;
        for(int i = 0; i<len1; i++) {
            for(int j = 0; j<len2; j++) {
                if(dp[i][j] > mx) {
                    mx = dp[i][j];
                    ans = str1.substr(i-mx+1, mx);
                }
            }
        }
        return ans;
    }
};