题解 | #最长公共子序列(二)#

#include <functional>
#include <vector>
class Solution {
public:
    /**
     * longest common subsequence
     * @param s1 string字符串 the string
     * @param s2 string字符串 the string
     * @return string字符串
     */
    string LCS(string s1, string s2) {
        int n1 = s1.length();
        int n2 = s2.length();
        vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1));
        vector<vector<int>> b(n1 + 1, vector<int>(n2 + 1));
        for (int i = 1; i <= n1; ++i) {
            for (int j = 1; j <= n2; ++j) {
                if (s1[i - 1] == s2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    b[i][j] = 1;
                } else if (dp[i - 1][j] > dp[i][j - 1]) {
                    dp[i][j] = dp[i - 1][j];
                    b[i][j] = 2;
                } else {
                    dp[i][j] = dp[i][j - 1];
                    b[i][j] = 3;
                }
            }
        }
        function<string(int, int)> getres = [&](int i, int j) -> string {
            if (i == 0 || j == 0) {
                return "";
            }
            string res;
            switch (b[i][j]) {
                case 1: {
                    res = getres(i - 1, j - 1) + s1[i - 1];
                } break;
                case 2: {
                    res = getres(i - 1, j);
                } break;
                case 3: {
                    res = getres(i, j - 1);
                } break;
            }
            return res;
        };
        return getres(n1, n2) == ""? "-1": getres(n1, n2);
    }
};

思路：动态规划。

在求最长公共子序列时，记录子序列扩展的方向，以在找到公共子序列最长的长度后，反向求出该子序列。