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161EditDistance.cs
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161EditDistance.cs
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using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace EditDistance
{
class Program
{
static void Main(string[] args)
{
int result1 = minDistance("abc", "abcd");
int result2 = minDistance("abc", "bca");
int result3 = minDistance("abc", "cba");
}
/**
* Latest update: June 17, 2015
* Leetcode: edit distance
* DP bottom-up
* Dynamic programming
* Time O(mxn), space O(mxn),
*
* http://blog.csdn.net/fightforyourdream/article/details/13169573
* there is a table to explain how to build this distance table:
* http://www.cnblogs.com/etcow/archive/2012/08/30/2662985.html
* Test cases:
* 1. abc, abcd, minimum distance is 1
* 2. abc, bca, minimum distance is 2
* 3. abc, cba, minimum distance is 2
* 4.
* Leetcode online judge:
* 1146 / 1146 test cases passed.
Status: Accepted
Runtime: 208 ms
* */
public static int minDistance(String word1, String word2) {
int l1 = word1.Length;
int l2 = word2.Length;
int[][] distance = new int[l1+1][];
for(int i = 0; i<l1+1;i++)
distance[i] = new int[l2+1];
// 边界情况:当其中一个string为空时,只要一直添加或删除就可以
for(int i=0; i< l1+1; i++){
distance[i][0] = i;
}
for(int j=1; j< l2+1; j++){
distance[0][j] = j;
}
// 递推,[i][j]处可以由左,上,左上3种情况而来
for(int i=1; i< l1+1; i++){
for(int j=1; j< l2+1; j++){
int tmp = Math.Min(distance[i-1][j]+1, // 从上演变
distance[i][j-1]+1); // 从左演变
// 从左上演变,考虑是否需要替换
// avoid word1, word2 index out of range bug - how?
// word1: index range from 0->l1-1
// word2: index range from 0->l2-1
// but distance[i][j] is between word1's substring from 0 to i-1 and word2's substring from i to j-1;
// not 0 to i.
//
distance[i][j] = Math.Min(tmp, distance[i-1][j-1]+((word1[i-1]==word2[j-1]) ? 0 : 1));
}
}
// 返回右下角
return distance[l1][l2];
}
}
}