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Edit_Distance.cpp
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Edit_Distance.cpp
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/*
Author: Weixian Zhou, ideazwx@gmail.com
Date: June 28, 2012
Problem: Edit Distance
Difficulty: medium
Source: http://www.leetcode.com/onlinejudge
Notes:
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
Solution:
This is a dynamic programming problem. The problem is similar to longest common
substring problem. Here we define f[n, m] as the minimum edit distance of string
a[0..n] and b[0..m]. Then
f[n, m] = f[n-1,m-1], if a[n] == b[m]
= min{f[n-1,m]+1, f[n,m-1]+1, f[n-1,m-1]+1}, if a[n] != b[m].
*/
#include <vector>
#include <set>
#include <climits>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
using namespace std;
class Solution {
public:
int minimal(int a, int b, int c) {
int min = a;
if (b < min) min = b;
if (c < min) min = c;
return min;
}
int minDistance(string word1, string word2) {
int ed[word1.length() + 1][word2.length() + 1];
// initialize
for (int i = 0; i < word1.length() + 1; i++) {
ed[i][0] = i;
}
for (int i = 0; i < word2.length() + 1; i++) {
ed[0][i] = i;
}
// DP
for (int i = 1; i < word1.length() + 1; i++) {
for (int j = 1; j < word2.length() + 1; j++) {
if (word1[i - 1] == word2[j - 1]) {
ed[i][j] = ed[i -1][j - 1];
}
else {
ed[i][j] = minimal(ed[i - 1][j] + 1,
ed[i][j - 1] + 1,
ed[i - 1][j - 1] + 1);
}
}
}
return ed[word1.length()][word2.length()];
}
};