题解 | 最长公共子序列(二)

class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 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) {
        // write code here
        int len1 = s1.length(), len2 = s2.length();
        string ans;
        if (len1 == 0 || len2 == 0) {
            return "-1";
        }
        vector<vector<int>>dp(len1 + 1, vector<int>(len2 + 1, 0));
        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                dp[i][j] = 0;
            }
        }
        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                if (s1[i-1] == s2[j-1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        int a = len1, b = len2;
        while(dp[a][b]>0){
            if(dp[a][b]>dp[a][b-1]&&dp[a][b]>dp[a-1][b]){
                ans += s1[a-1];
                b--;
                a--;
                continue;
            }
            if(dp[a][b] == dp[a][b-1]){
                b--;
                continue;
            }
            if(dp[a][b] == dp[a-1][b]){
                a--;
                continue;
            }
        }
        reverse(ans.begin(),ans.end());
        if(ans.length()<1){
            return "-1";
        }

        return ans;
    }
};