Find the longest common sub-sequence

Description:

Given 3 strings of all having length < 100,the task is to find the longest common sub-sequence
in all three given sequences.

Example:

Input : str1 = "geeks"
str2 = "geeksfor"
str3 = "geeksforgeeks"
Output : 5
Longest common subsequence is "geeks"
i.e., length = 5
Input : str1 = "abcd1e2"
str2 = "bc12ea"
str3 = "bd1ea"
Output : 3
Longest common subsequence is "b1e"
i.e. length = 3.

解题方法：

在两个string中寻找LCS:
假设string1的长度为m，string2的长度为n。
则我们创建一个m+1 * n+1的矩阵，矩阵DP[i][j]代表的是，当取string1前i个字符和取string2前j个字符时，它们的LCS长度为多少。
则状态转移方程为：

当string1[i - 1] == string2[j - 1]，则DP[i][j] = DP[i - 1][j - 1] + 1；这代表着当我们从string1和string2再各自取出一个字符，并且这两个字符相同，这两个substring的LCS的长+=1。
否则，DP[i][j] = max(DP[i - 1][j], DP[i][j - 1]);

这道题是寻找两个string的LCS的升级版。
状态转移方程为：

当string1[x - 1] == string2[y - 1] == string3[z - 1]，则DP[x][y][z] = DP[x - 1][y - 1][z - 1] + 1；
否则，DP[x][y][z] = max(DP[x - 1][y][z], DP[x][y - 1][z], DP[x][y][z - 1]);

Time Complexity：

O(n^3)

完整代码：

int longestSubSequence(string& str1, string& str2, string& str3) {
    int size1 = str1.size(), size2 = str2.size(), size3 = str3.size();
    vector<vector<vector<int>>> DP(size1 + 1, vector<vector<int>>(size2 + 1, vector<int>(size3 + 1, 0)));
    for(int x = 1; x <= size1; x++) {
        for(int y = 1; y <= size2; y++) {
            for(int z = 1; z <= size3; z++) {
                if(str1[x - 1] == str2[y - 1] && str2[y - 1] == str3[z - 1]) {
                    cout<<str1[x]<<endl;
                    DP[x][y][z] = DP[x - 1][y - 1][z - 1] + 1;
                }   
            else
                    DP[x][y][z] = max(DP[x - 1][y][z], max(DP[x][y -1][z], DP[x][y][z - 1]));
            }
        }
    }
    return DP[size1][size2][size3];
}