128 lines
3.8 KiB
C++
128 lines
3.8 KiB
C++
// Source : https://oj.leetcode.com/problems/edit-distance/
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// Author : Hao Chen
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// Date : 2014-08-22
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/**********************************************************************************
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*
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* Given two words word1 and word2, find the minimum number of steps required to
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* convert word1 to word2. (each operation is counted as 1 step.)
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*
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* You have the following 3 operations permitted on a word:
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*
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* a) Insert a character
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* b) Delete a character
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* c) Replace a character
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*
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*
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**********************************************************************************/
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#include <iostream>
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#include <string>
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#include <vector>
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#include <algorithm>
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using namespace std;
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/*
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* Dynamic Programming
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*
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* Definitaion
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*
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* m[i][j] is minimal distance from word1[0..i] to word2[0..j]
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*
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* So,
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*
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* 1) if word1[i] == word2[j], then m[i][j] == m[i-1][j-1].
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*
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* 2) if word1[i] != word2[j], then we need to find which one below is minimal:
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*
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* min( m[i-1][j-1], m[i-1][j], m[i][j-1] )
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*
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* and +1 - current char need be changed.
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*
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* Let's take a look m[1][2] : "a" => "ab"
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*
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* +---+ +---+
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* ''=> a | 1 | | 2 | '' => ab
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* +---+ +---+
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*
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* +---+ +---+
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* a => a | 0 | | 1 | a => ab
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* +---+ +---+
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*
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* To know the minimal distance `a => ab`, we can get it from one of the following cases:
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*
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* 1) delete the last char in word1, minDistance( '' => ab ) + 1
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* 2) delete the last char in word2, minDistance( a => a ) + 1
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* 3) change the last char, minDistance( '' => a ) + 1
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*
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*
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* For Example:
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*
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* word1="abb", word2="abccb"
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*
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* 1) Initialize the DP matrix as below:
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*
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* "" a b c c b
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* "" 0 1 2 3 4 5
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* a 1
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* b 2
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* b 3
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*
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* 2) Dynamic Programming
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*
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* "" a b c c b
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* "" 0 1 2 3 4 5
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* a 1 0 1 2 3 4
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* b 2 1 0 1 2 3
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* b 3 2 1 1 2 2
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*
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*/
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int min(int x, int y, int z) {
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return std::min(x, std::min(y,z));
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}
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int minDistance(string word1, string word2) {
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int n1 = word1.size();
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int n2 = word2.size();
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if (n1==0) return n2;
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if (n2==0) return n1;
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vector< vector<int> > m(n1+1, vector<int>(n2+1));
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for(int i=0; i<m.size(); i++){
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m[i][0] = i;
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}
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for (int i=0; i<m[0].size(); i++) {
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m[0][i]=i;
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}
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//Dynamic Programming
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int row, col;
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for (row=1; row<m.size(); row++) {
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for(col=1; col<m[row].size(); col++){
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if (word1[row-1] == word2[col-1] ){
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m[row][col] = m[row-1][col-1];
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}else{
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int minValue = min(m[row-1][col-1], m[row-1][col], m[row][col-1]);
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m[row][col] = minValue + 1;
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}
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}
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}
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return m[row-1][col-1];
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}
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int main(int argc, char**argv)
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{
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string word1="abb", word2="abccb";
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if (argc>2){
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word1 = argv[1];
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word2 = argv[2];
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}
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int steps = minDistance(word1, word2);
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cout << word1 << ", " << word2 << " : " << steps << endl;
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return 0;
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}
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