百度-20120926-字符串编辑距离

来源：互联网发布：什么软件可以修复ie 编辑：程序博客网时间：2024/06/06 12:53

题目：找出字符串的编辑距离，即统计一个字符串s1最少经过多少步操作变成字符串s2？（操作有三种：添加一个字符，删除一个字符，修改一个字符）。

备注：1）一直都推荐阅读最精彩的参考资料，关于Levenshtein distance也即edit distance，请参考wiki：http://en.wikipedia.org/wiki/Levenshtein_distance，wiki给出了非常详细的解答，相信读过之后会让你有收获。2）这题出现了很多大公司的笔试/面试题中，比如google，百度 and so on.

// 这是递归的解法，空间复杂度相当高，不实用// calculate the edit distance between source and target recursivelyint editDistanceRecursive(string source, string target){int len_source = source.length();int len_target = target.length();// test for degenerate cases of empty string of source or target stringif(len_source == 0)return len_target;if(len_target == 0)return len_source;int cost = 0;// check the last character of source and target is equal or notif(source[len_source-1] != target[len_target-1])cost = 1;return minimum(editDistanceRecursive(source.substr(0, len_source-1), target)+1, editDistanceRecursive(source, target.substr(0, len_target-1))+1, editDistanceRecursive(source.substr(0, len_source-1), target.substr(0, len_target-1))+cost);}

// 面试中应该倾向采用非递归的解法int editDistanceIteratively(string source, string target){int len_source = source.length();int len_target = target.length();// test for degenerate cases of empty string of source or target stringif(len_source == 0)return len_target;if(len_target == 0)return len_source;// do initilizationint** dMatrix = new int*[len_source+1];for(int i = 0; i <= len_source; i++){dMatrix[i] = new int[len_target+1];memset(dMatrix[i], 0, sizeof(int)*(len_target+1));}for(int i = 1; i <= len_source; i++)dMatrix[i][0] = i;for(int i = 1; i <= len_target; i++)dMatrix[0][i] = i;for(int j = 1; j <= len_target; j++)for(int i = 1; i <= len_source; i++){if(source[i-1] == target[j-1]){dMatrix[i][j] = dMatrix[i-1][j-1];}else{dMatrix[i][j] = minimum(dMatrix[i-1][j]+1, dMatrix[i][j-1]+1, dMatrix[i-1][j-1]+1);}}return dMatrix[len_source][len_target];}