动态规划 - 求最长公共子序列

    注意：区分下字串和子序列，字串是连续的，子序列是原序列任意抽出若干元素的结果。

    

    int longestCommonSubsequence(string text1, string text2) {
        //金典的动态规划题目!
        //状态定义：dp[i][j] 表示 s1[0..i-1], s2[0..j-1] 的最长 LCS; i表示s1的元素个数
        //base case：dp[0][j] = dp[i][0] = 0
        //状态转移，如何根据 dp[i-1][j-1] 求出 dp[i][j] ? 选择的关键是 s[i] ?= s[j]
        //如果新加入的两个字符 s[i] == s[j]，那么 dp[i][j] = dp[i-1][j-1] + 1
        //如果新加入的两个字符 s[i] != s[j]，那么LCS要么包含s[i]，要么包含s[j]，要么都不包含：
        //dp[i][j] = max(dp[i-1][j-1], dp[i-1][j], dp[i][j-1])
        string lcs;
        if(text1.empty() || text2.empty()) return 0;
        int n1 = text1.size(), n2 = text2.size();
        vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1, 0));
        vector<vector<char>> dpC(n1 + 1, vector<char>(n2 + 1, ' ')); //求具体的LCS值
        for(int i = 1; i <= n1; ++i){
            for(int j = 1; j <= n2; ++j){
                if(text1[i-1] == text2[j-1]){
                    dp[i][j] = dp[i-1][j-1] + 1;
                    dpC[i][j] = '\\'; //往左上方
                    //尝试获取具体结果
                }else{
                    dp[i][j] = max(dp[i][j-1], dp[i-1][j]);
                    //注意，根据dp的定义，因为 dp[i-1][j-1] 一定小于 dp[i-1][j], dp[i][j-1]
                    //所以这里也就不需要 dp[i-1][j-1] 参与求最大值了
                    if(dp[i][j] == dp[i][j-1]) dpC[i][j] = '-'; //往左边
                    else dpC[i][j] = '|'; //往上面
                }
            }
        }

        //调试信息
        printf("s1:%s\n", text1.c_str());
        printf("s2:%s\n", text2.c_str());

        printf("\n");
        printf("  0- ");
        int k = 0;
        for(auto c:text2){
            printf("%d%c ", ++k, c);
        }
        printf("\n"); k = 0;
        for(int i = 0; i <= n1; ++i){
            if(i == 0) printf("0- ");
            else printf("%d%c ",++k, text1[i - 1]);
            for(int j = 0; j <= n2; ++j){
                printf("%d%c  ", dp[i][j], dpC[i][j]);
            }
            printf("\n");
        }
        printf("----------------------------\n");

        //获取LCS的值，思路，遇到 '-' 往左走；遇到 '|' 往上走；遇到 '\' 打印该值
        int i = n1, j = n2;
        while(i > 0 && j > 0){
            if(dpC[i][j] == '-') --j;
            else if(dpC[i][j] == '|') --i;
            else if(dpC[i][j] == '\\'){
                lcs.push_back(text1[i - 1]);
                printf("%c %d:%d\n", text1[i - 1], i, j);
                --i; --j;
            }
        }
        printf("----------------------------\n");
        printf("s1:%s\n", text1.c_str());
        printf("s2:%s\n", text2.c_str());
        reverse(lcs.begin(), lcs.end());
        printf("lcs %d:%s\n", lcs.size(), lcs.c_str());

        return dp[n1][n2];
    }

    参考：《算法导论》