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 hereint 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 hereint 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;}
};