[LeetCode][Java] Edit Distance

题目：

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

题意：

给定两个单词word1 and word2，找出从word1变为word2
所需要的最少的步数.（每次操作被看为一步.）

对于每个单词你能够有如下的三种操作：

a) 插入一个字符

b) 删去一个字符

c) 替换一个字符

算法分析：

动态规划

初始情况: dp[i][0] = i (0 <= i <= word1 length), dp[0][j] = j (0 <= j <= word2 length).

初始化的时候我们是在第0行，也就是另一个字符串一个字符都不取，即空字符串。 res[i]表示如何将当前字符串（前i个元素）变成一个空字符串， 所以我们可以进行i次删除（或者插入）操作，就可以使得当前字符串变成空字符串（或者空字符串变成当前字符串）

* dp[i][j]表示word1前i个字母变成word2前j个字母的步数(edit distance)。

*

* 如果word1的第i个字母等于word2的第j个字母, 则dp[i][j] = dp[i-1][j-1]。如果不等, 则有三种情况:

*

* 1) 把word1的前i-1个字母变成word2的前j-1个字母, 再把word1的第i个字母换成word2的第j个字母, 即dp[i-1][j-1] + 1

*

* 2) 把word1的前i个字母变成word2的前j-1个字母, 再加上word2的第j个字母, 即dp[i][j-1] + 1

*

* 3) 删掉word1的第i个字母, 把word1的前i-1个字母变成word2的前j个字母, 即1 + dp[i-1][j]

AC代码：

public class Solution
{
public static int minDistance(String word1, String word2)
{
int len1 = word1.length();
int len2 = word2.length();

// len1+1, len2+1, because finally return dp[len1][len2]
int[][] dp = new int[len1 + 1][len2 + 1];

for (int i = 0; i <= len1; i++)
dp[i][0] = i;

for (int j = 0; j <= len2; j++)
dp[0][j] = j;

//iterate though, and check last char
for (int i = 1; i <= len1; i++)
{
char c1 = word1.charAt(i-1);
for (int j = 1; j <= len2; j++)
{
char c2 = word2.charAt(j-1);

//if last two chars equal
if (c1 == c2)
{
//update dp value for +1 length
dp[i][j] = dp[i-1][j-1];
}
else
{
//把word1的前i-1个字母变成word2的前j-1个字母, 再把word1的第i个字母换成word2的第j个字母, 即dp[i-1][j-1] + 1
int replace = dp[i-1][j-1] + 1;
int delete = 1 + dp[i-1][j] ;//删掉word1的第i个字母, 把word1的前i-1个字母变成word2的前j个字母, 即1 + dp[i-1][j]
int insert= dp[i][j-1] + 1;//把word1的前i个字母变成word2的前j-1个字母, 再加上word2的第j个字母, 即dp[i][j-1] + 1

int min = Math.min(replace, insert);
min = Math.min(min,delete);
dp[i][j] = min;
}
}
}
return dp[len1][len2];
}
}

参考:
http://www.cnblogs.com/springfor/p/3896167.html?utm_source=tuicool http://blog.csdn.net/linhuanmars/article/details/24213795 http://chaoren.is-programmer.com/posts/44402.html