Edit Distance 字符串距离（重重）

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题目：

链接

解答：

编程之美上有解答，如果采用递归的方法，会超时。

解答一，采用递归，代码：

  class Solution {  public:  int minDistance(string word1, string word2) {  return calculate(word1, word2);  }  int calculate(string word1, string word2)  {  if (word1 == "")  {  if (word2 == "")  return 0;  else  return word2.length();  }  if (word2 == "")  {  if (word1 == "")  return 0;  else  return word1.length();  }  if (word1[0] == word2[0])  {  string w1 = word1.substr(1, word1.length() - 1);  string w2 = word2.substr(1, word2.length() - 1);  return calculate(w1, w2);  }  else  {  string w1 = word1.substr(1, word1.length() - 1);  string w2 = word2.substr(1, word2.length() - 1);  int t1 = calculate(w1, word2);  int t2 = calculate(word1, w2);  int t3 = calculate(w1, w2);  int minval = min(t1, t2);  minval = min(t3, minval);  return minval + 1;  }  }  };

解答二：

动态规划，http://www.cnblogs.com/lihaozy/archive/2012/12/31/2840152.html。

代码：

  class Solution {  public:  int minDistance(string word1, string word2) {  int row = word1.length() + 1;  int col = word2.length() + 1;  vector<vector<int> > f(row, vector<int>(col));  for (int i = 0; i < row; i++)  f[i][0] = i;  for (int i = 0; i < col; i++)  f[0][i] = i;  for (int i = 1; i < row; i++)  {  for (int j = 1; j < col; j++)  {  if (word1[i-1] == word2[j-1])  f[i][j] = f[i - 1][j - 1];  else  {  f[i][j] = f[i - 1][j - 1] + 1;  f[i][j] = min(f[i][j], min(f[i - 1][j] + 1, f[i][j - 1] + 1));  }  }  }  return f[row - 1][col - 1];  }  };

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