LintCode Longest Common Subsequence(最长公共子序列长度，动态规划入门题)

#include <iostream>
#include <cstring>

using namespace std;

class Solution {
public:
/**
* @param A, B: Two strings.
* @return: The length of longest common subsequence of A and B.
*/
int longestCommonSubsequence(string A, string B) {
// write your code here
//得出两个字符串的长度，如果长度为0直接返回0
int lenA = A.length();
int lenB = B.length();
if(lenA == 0 && lenA == lenB)
return 0;
//再字符串头添加空位，方便之后编码从1开始
A = " " + A;
B = " " + B;
int chess[lenA + 1][lenB + 1];
//将表中所有元素置为0
memset(chess, 0, sizeof(int) * lenA * lenB);
for (int i = 1; i <= lenA; i++) {
for (int j = 1; j <= lenB; j++) {
//当两个字符串中 A[i] == A[j],
// 则最长LCS = LCS{（A[i - 1]， A[j- 1]）} + 1
//否则就在LCS{（A[i - 1]， A[j]）}，和 LCS{（A[i]， A[j- 1]）}中选最长的那个
//这里的A[i]，A[j]均指的是字符串
if (A[i] == B[j]) {
chess[i][j] = chess[i - 1][j - 1] + 1;
} else {
chess[i][j] = max(chess[i - 1][j], chess[i][j - 1]);
}
}
}
return chess[lenA][lenB];
}
};

int main() {

Solution slo;
string str1 = "";
string str2 = "";
cout << slo.longestCommonSubsequence(str1, str2) << endl;
return 0;
}