Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)
You have the following 3 operations permitted on a word:
a) Insert a character
b) Delete a character
c) Replace a character
思路I：带回溯的递归。但结果Time Limit Exceeded.
class Solution { public: int minDistance(string word1, string word2) { // Start typing your C/C++ solution below // DO NOT write int main() function if(word1.empty()) return word2.length(); else if(word2.empty()) return word1.length(); result = max(word1.length(),word2.length()); while(word1[0] == word2[0]){ if(word2.length()==1) { result = word1.length()-1; return result; } else if(word1.length()==1) { result = word2.length()-1; return result; } word1 =word1.substr(1,word1.length()-1); word2 = word2.substr(1,word2.length()-1); } replaceOpr(word1,word2,1); //replace一般所用的操作最少，所以写在最前面 insertOpr(word1,word2,1); deleteOpr(word1,word2,1); return result; } void insertOpr(string word1, string word2, int oprNum) { if(word2.length()==1) { oprNum += word1.length(); if(oprNum < result) result = oprNum; return; } word2 = word2.substr(1,word2.length()-1); //insert while(word1[0] == word2[0]){ if(word2.length()==1) { oprNum += word1.length()-1; if(oprNum < result) result = oprNum; return; } else if(word1.length()==1) { oprNum += word2.length()-1; if(oprNum < result) result = oprNum; return; } word1 =word1.substr(1,word1.length()-1); word2 = word2.substr(1,word2.length()-1); } if(oprNum+1>result) return; replaceOpr(word1,word2,oprNum+1); insertOpr(word1,word2, oprNum+1); deleteOpr(word1,word2,oprNum+1); } void deleteOpr(string word1, string word2, int oprNum) { if(word1.length() == 1) { oprNum += word2.length()-1; if(oprNum < result) result = oprNum; return; } word1 =word1.substr(1,word1.length()-1); //delete while(word1[0] == word2[0]){ if(word2.length()==1) { oprNum += word1.length()-1; if(oprNum < result) result = oprNum; return; } else if(word1.length()==1) { oprNum += word2.length()-1; if(oprNum < result) result = oprNum; return; } word1 =word1.substr(1,word1.length()-1); word2 = word2.substr(1,word2.length()-1); } if(oprNum+1>result) return; replaceOpr(word1,word2,oprNum+1); insertOpr(word1,word2, oprNum+1); deleteOpr(word1,word2,oprNum+1); } void replaceOpr(string word1, string word2, int oprNum) { if(word2.length()==1) { oprNum += word1.length()-1; if(oprNum < result) result = oprNum; return; } else if(word1.length() == 1) { oprNum += word2.length()-1; if(oprNum < result) result = oprNum; return; } word1 =word1.substr(1,word1.length()-1); word2 = word2.substr(1,word2.length()-1); while(word1[0] == word2[0]){ if(word2.length()==1) { oprNum += word1.length()-1; if(oprNum < result) result = oprNum; return; } else if(word1.length()==1) { oprNum += word2.length()-1; if(oprNum < result) result = oprNum; return; } word1 =word1.substr(1,word1.length()-1); word2 = word2.substr(1,word2.length()-1); } if(oprNum+1>result) return; replaceOpr(word1,word2,oprNum+1); insertOpr(word1,word2, oprNum+1); deleteOpr(word1,word2, oprNum+1); } private: int result; };
思路II：动态规划。将所要求的min step作为状态，dp[i][j]表示word2的前j各字符通过word1的前i各字符转换最少需要多少步。可以看到有两个以上的string，通常状态要定义为二维数组，表示两个字符串前几个字符之间的关系。
class Solution { public: int minDistance(string word1, string word2) { vector<vector<int> > f(word1.size()+1, vector<int>(word2.size()+1)); f[0][0] = 0; for(int i = 1; i <= word2.size(); i++) f[0][i] = i; for(int i = 1; i <= word1.size(); i++) f[i][0] = i; for(int i = 1; i <= word1.size(); i++) for(int j = 1; j <= word2.size(); j++) { f[i][j] = INT_MAX; if (word1[i-1] == word2[j-1]) f[i][j] = f[i-1][j-1]; f[i][j] = min(f[i][j], f[i-1][j-1] + 1); //replace f[i][j] = min(f[i][j], min(f[i-1][j], f[i][j-1]) + 1); //delete or insert } return f[word1.size()][word2.size()]; } };
72. Edit Distance (String; DP)
原文：http://www.cnblogs.com/qionglouyuyu/p/4920976.html