标签：
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:
For example, given word1 = "mart" and word2 = "karma", return 3.
DP 四要素：1. State 2. Function 3. Initialization 4. Answer
令dp[i][j]表示长度为i的word1和长度为j的word2的最小距离。假设末位分别为x和y。那么根据x和y是否相同，考虑情况如下：
1. x == y
dp[i][j] = dp[i-1][j-1]
2. x != y
a) Delete x => dp[i-1][j] + 1
b) Insert y => dp[i][j-1] + 1
c) Replace x with y => dp[i-1][j-1] + 1
dp[i][j]取a,b,c中最小值。
public class Solution {
/**
* @param word1 & word2: Two string.
* @return: The minimum number of steps.
*/
public int minDistance(String word1, String word2) {
// write your code here
int len1 = word1.length();
int len2 = word2.length();
// dp[i][j] represents min distance for word1[0, i - 1] and word2[0, j - 1]
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;
}
for (int i = 1; i <= len1; i++) {
for (int j = 1; j <= len2; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
// delete word1[i - 1]
int x = dp[i - 1][j] + 1;
// insert word2[j - 1] into word1
int y = dp[i][j - 1] + 1;
// replace word1[i - 1] with word2[j - 1]
int z = dp[i - 1][j - 1] + 1;
dp[i][j] = Math.min(x, y);
dp[i][j] = Math.min(dp[i][j], z);
}
}
}
return dp[len1][len2];
}
}
标签：
原文地址：http://www.cnblogs.com/ireneyanglan/p/5980439.html
