lintcode - 删除数字
1 class Solution { 2 public: 3 /* 4 * @param A: A positive integer which has N digits, A is a string 5 * @param l: Remove k digits 6 * @return: A string 7 */ 8 string dp[220][220]; 9 string inf; 10 string DeleteDigits(string &A, int l) { 11 // write your code here 12 if(A.size() == 0) return ""; 13 if(l == 0) return A; 14 int len = A.length() - l; 15 for(int i = 0; i < A.size(); ++i){ 16 dp[i][0] = ""; 17 inf = inf + "9"; 18 } 19 for(int i = 0; i < A.size(); ++i){ 20 for(int j = 0; j < A.size(); ++j){ 21 dp[i][j] = inf; 22 } 23 } 24 vector<string> vec; 25 for(int i = 0; i < A.length(); ++i){ 26 stringstream ss; 27 ss << (A[i] - '0'); 28 vec.push_back(ss.str()); 29 } 30 dp[0][1] = A.substr(0,1); 31 for(int i = 1; i < vec.size(); ++i){ 32 for(int k = 1; k <= i + 1 && k <= len; ++k){ 33 for(int j = 0; j < i; ++j){ 34 if(k <= j + 1) 35 dp[i][k] = min(dp[j][k - 1] + vec[i], min(dp[i][k], dp[j][k])); 36 else dp[i][k] = min(dp[j][k - 1] + vec[i], dp[i][k]); 37 38 if(k == 1) dp[i][k] = min(dp[i][k], vec[i]); 39 } 40 // cout << dp[2][1] << endl; 41 } 42 43 } 44 int len1 = dp[vec.size() - 1][len].length(); 45 for(int i = 0; i < len1; ++i){ 46 if(dp[vec.size() - 1][len][i] != '0'){ 47 dp[vec.size() - 1][len] = dp[vec.size() - 1][len].substr(i,len1-i); 48 break; 49 } 50 } 51 return dp[vec.size() - 1][len]; 52 53 } 54 };
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
