lintcode - 删除数字

 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[][];
      string inf;
     string DeleteDigits(string &A, int l) {
         // write your code here
         if(A.size() == ) return "";
         if(l == ) return A;
         int len = A.length() - l;
         for(int i = ; i < A.size(); ++i){
                 dp[i][] = "";
                 inf = inf + "";
         }
         for(int i = ; i < A.size(); ++i){
                 for(int j = ; j < A.size(); ++j){
                         dp[i][j] = inf;
                 }
         }
         vector<string> vec;
         for(int i = ; i < A.length(); ++i){
                 stringstream ss;
                 ss << (A[i] - '');
                 vec.push_back(ss.str());
         }
         dp[][] = A.substr(,);
         for(int i = ; i < vec.size(); ++i){
                         for(int k = ; k <= i +  && k <= len; ++k){
                                 for(int j = ; j < i; ++j){
                                         if(k <= j + )
                                         dp[i][k] = min(dp[j][k - ] + vec[i], min(dp[i][k], dp[j][k]));
                                         else dp[i][k] = min(dp[j][k - ] + vec[i], dp[i][k]);

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

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