Edit distance problem 动态规划和递归解法之比较

问题：

给定字符串firstStr，secondStr，长度分别为m，n。通过delete, insert, replace 操作把firstStr转化为secondStr,而这三种操作的每次执行成本分别为delCost, insCost, repCost。求完成这种转换的最小成本。

解决方案：

1. 递归（recursive)

base condition：用于终止递归， 当m == 0 || n == 0;

Mincost( firstStr, secondStr, m, n, delCost, insCost, repCost ) = 

      min{ Mincost( firstStr, secondStr, m - 1, n, delCost, insCost, repCost ) + delCost,                                                                                                                                                         Mincost( firstStr, secondStr, m, n-1, delCost, insCost, repCost )+ insCost,

                 Mincost( firstStr, secondStr, m - 1, n-1, delCost, insCost, repCost)+ repCost };

2.动态规划（dynamic programming)

引入成本数组cost[m][n]：代表把长度m的字符串转成长度为n的字符串的成本；

建立递推关系：cost[m][n] = min{ cost[m-1][n] + delCost, cost[m][n-1] + insCost, cost[m-1][n-1] + repCost }；

两种方法的比较的区别很明显：

1.递归计算整个搜索过程最终会形成一棵树，这颗树不同树枝之间会有很多重复的计算，最终导致了性能很差；

  2.动态规划会把许多计算结果缓存到cost数组里，虽然有个空间的开销，但是算过的不用再计算，直接查表即可，

       所以动态规划算法有很好的性能表现

代码如下：

#ifndef _EDIT_DISTANCE_H_#define _EDIT_DISTANCE_H_#include <string.h>#include <stdlib.h>#include <stdio.h>/**helper function**/int minVal( int a, int b ){return ( a < b ) ? a : b;}/**helper function**/inline int minnum( int a, int b, int c ){return minVal( minVal(a, b), c );}/**implementation of recursive for edit distance problem**/int EditDistanceRecur( const char* firstStr, const char* secondStr, int firstPos, int secondPos, int delCost, int insCost, int rapCost ){if( 0 == firstPos && 0 == secondPos )return 0;if( 0 == firstPos )return secondPos;if( 0 == secondPos )return firstPos;int del = EditDistanceRecur( firstStr, secondStr, firstPos - 1, secondPos, delCost, insCost, rapCost ) + delCost;int insert = EditDistanceRecur( firstStr, secondStr, firstPos, secondPos - 1, delCost, insCost, rapCost  ) + insCost;int replace = EditDistanceRecur( firstStr, secondStr, firstPos - 1, secondPos - 1, delCost, insCost, rapCost  );if( firstStr[firstPos - 1] != secondStr[secondPos - 1] )replace += rapCost;return minnum( del, insert, replace );}/** interface function **/int FindEditDistance( const char* strfirst, const char* strsecond, int delCost, int insCost, int rapCost ){return EditDistanceRecur( strfirst, strsecond, strlen( strfirst ), strlen( strsecond ), delCost, insCost, rapCost );}/** implementation of dynamic programming for edit distance problem**/int EditDistanceDP( const char* firstStr, const char* secondStr, int delCost, int insCost, int rapCost ){size_t firstLen = strlen( firstStr );size_t secondLen = strlen( secondStr );int** table = new int*[ firstLen + 1];for( int i = 0; i <= firstLen; i++ ){table[i] = new int[ secondLen + 1];memset( table[i], 0x00, sizeof(int)*( secondLen + 1 ) );}//base conditionfor( int i = 0; i <= secondLen; i++ ){table[0][i] = i * insCost;}for( int i = 0; i <= firstLen; i++ ){table[i][0] = i * delCost;}for( int i = 1; i <= firstLen; i++ ){for( int j = 1; j <= secondLen; j++ ){int deleteCost = table[i][j-1];deleteCost += delCost;int insertCost = table[i-1][j];insertCost += insCost;int replaceCost = table[i-1][j-1];if( firstStr[i] != secondStr[j] )replaceCost += rapCost;table[i][j] = minnum( deleteCost, insertCost, replaceCost );}}int res = table[firstLen][secondLen];for( int i = 0; i <= firstLen; i++ ){delete [] table[i];}if( table ){delete [] table;}return res;}/** test method**/void TestEditDistance(){const char* first = "adddfooiedfafa";const char* second = "adafsadoiddf";int delCost = 1;int insCost = 2;int rapCost = 3;printf("Minimum cost  %d by dynamic programming \n",    EditDistanceDP( first, second, delCost, insCost, rapCost ) );printf("Minimum cost  %d by recursive \n",first, second, FindEditDistance( first, second, delCost, insCost, rapCost ) );getchar();}#endif