EditDistance

Change-Id: I82e3cd5a3bb94a55d55aa006d7d5c82932814b47

Author: applewjg

EditDistance.java

@@ -0,0 +1,63 @@
+/*
+ Author:     King, [email protected]
+ Date:       Dec 25, 2014
+ Problem:    Edit Distance
+ Difficulty: Medium
+ Source:     https://oj.leetcode.com/problems/edit-distance/
+ 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: Dynamic Programming.
+           1. Time: O(mn) Space: O(mn)
+           2. Time: O(mn) Space: O(n);
+ */
+public class Solution {
+    public int minDistance_1(String word1, String word2) {
+        if(word1==word2) return 0;
+        int len1 = word1.length();
+        int len2 = word2.length();
+        int[][] dp = new int[len1+1][len2+1];
+        
+        for(int i=0;i<=len1;i++)
+            dp[i][0] = i;
+        for(int i=0;i<=len2;i++)
+            dp[0][i] = i;
+        
+        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{
+                    dp[i][j] = Math.min(dp[i-1][j],Math.min(dp[i][j-1],dp[i-1][j-1]))+1;
+                }
+            }
+        }
+        return dp[len1][len2];
+    }
+    public int minDistance(String word1, String word2) {
+        if(word1==word2) return 0;
+        int len1 = word1.length();
+        int len2 = word2.length();
+        int[] dp = new int[len2+1];
+        
+        for(int i=0;i<=len2;i++)
+            dp[i] = i;
+        
+        for(int i=1;i<=len1;i++){
+            int upperLeftBak = dp[0];
+            dp[0] = i;
+            for(int j=1;j<=len2;j++){
+                int upperLeft = upperLeftBak;
+                upperLeftBak = dp[j];
+                if(word1.charAt(i-1)==word2.charAt(j-1)) dp[j] = upperLeft;
+                else{
+                    dp[j] = Math.min(dp[j],Math.min(dp[j-1],upperLeft))+1;
+                }
+            }
+        }
+        return dp[len2];
+    }
+}