lintcode

class Solution {
public:
    /*
     * @param A: A positive integer which has N digits, A is a string
     * @param l: Remove k digits
     * @return: A string
     */
     string dp[220][220];
     string inf;
    string DeleteDigits(string &A, int l) {
        // write your code here
        if(A.size() == 0) return "";
        if(l == 0) return A;
        int len = A.length() - l;
        for(int i = 0; i < A.size(); ++i){
                dp[i][0] = "";
                inf = inf + "9";
        }
        for(int i = 0; i < A.size(); ++i){
                for(int j = 0; j < A.size(); ++j){
                        dp[i][j] = inf;
                }
        }
        vector<string> vec;
        for(int i = 0; i < A.length(); ++i){
                stringstream ss;
                ss << (A[i] - '0');
                vec.push_back(ss.str());
        }
        dp[0][1] = A.substr(0,1);
        for(int i = 1; i < vec.size(); ++i){
                        for(int k = 1; k <= i + 1 && k <= len; ++k){
                                for(int j = 0; j < i; ++j){
                                        if(k <= j + 1)
                                        dp[i][k] = min(dp[j][k - 1] + vec[i], min(dp[i][k], dp[j][k]));
                                        else dp[i][k] = min(dp[j][k - 1] + vec[i], dp[i][k]);

                                        if(k == 1) dp[i][k] = min(dp[i][k], vec[i]);
                                }
                               // cout << dp[2][1] << endl;
                        }

        }
        int len1 = dp[vec.size() - 1][len].length();
        for(int i = 0; i < len1; ++i){
                if(dp[vec.size() - 1][len][i] != '0'){
                        dp[vec.size() - 1][len] = dp[vec.size() - 1][len].substr(i,len1-i);
                        break;
                }
        }
        return dp[vec.size() - 1][len];

    }
};

dp[i][j]代表以i结尾，长度为j的最小的字串 因为这个数字会很长 所以不能用数字来dp，必须用字符串来dp 注意一下因为是字符串 所以要去掉前导零