72. Edit Distance (JAVA)

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

1. Insert a character
2. Delete a character
3. Replace a character

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

 

使用递归会造成Time limit exceeded

class Solution {
    public int minDistance(String word1, String word2) {
        StringBuffer strBuf1 = new StringBuffer(word1);
        StringBuffer strBuf2 = new StringBuffer(word2);

        return dfs(strBuf1,strBuf2,0,0,0);
    }

    public int insert(StringBuffer strBuf1, StringBuffer strBuf2, int i1, int i2, int depth){
        strBuf1.insert(i1, strBuf2.charAt(i2));
        int ret = dfs(strBuf1,strBuf2,i1+1, i2+1,depth+1);
        strBuf1.deleteCharAt(i1); //recover
        return ret;
    }

    public int delete(StringBuffer strBuf1, StringBuffer strBuf2, int i1, int i2, int depth){
        Character ch = strBuf1.charAt(i1);
        strBuf1.deleteCharAt(i1);
        int ret = dfs(strBuf1,strBuf2,i1, i2,depth+1);
        strBuf1.insert(i1,ch); //recover;
        return ret;
    }

    public int replace(StringBuffer strBuf1, StringBuffer strBuf2, int i1, int i2, int depth){
        Character ch = strBuf1.charAt(i1);
        strBuf1.setCharAt(i1, strBuf2.charAt(i2));
        int ret = dfs(strBuf1,strBuf2,i1+1, i2+1,depth+1);
        strBuf1.setCharAt(i1, ch);
        return ret;
    }

    private int dfs(StringBuffer strBuf1, StringBuffer strBuf2, int i1, int i2, int depth){
        while(i1 < strBuf1.length() && i2 < strBuf2.length() && strBuf1.charAt(i1) == strBuf2.charAt(i2)){
            i1++;
            i2++;
        }

        if(i1 == strBuf1.length() && i2 == strBuf2.length()) return depth;
        if(i1 == strBuf1.length()) return depth+strBuf2.length()-i2;
        if(i2 == strBuf2.length()) return depth+strBuf1.length()-i1;

        int ret = insert(strBuf1,strBuf2,i1,i2,depth);
        ret = Math.min(ret,delete(strBuf1,strBuf2,i1,i2,depth));
        ret = Math.min(ret,replace(strBuf1,strBuf2,i1,i2,depth));

        return ret;

    }

}

使用动态规划dp[i][j]表示从word1[i+1]位置到word2[j+1]位置 需要改变次数。

class Solution {
    public int minDistance(String word1, String word2) {
        int[][] dp = new int[word1.length()+1][word2.length()+1];
        for(int i = 0; i <= word1.length(); i++){
            dp[i][0] = i;
        }
        for(int j = 0; j <= word2.length(); j++){
            dp[0][j] = j;
        }
        for(int i = 1; i <= word1.length(); i++){
            for(int j = 1; j <= word2.length(); j++){
                if(word1.charAt(i-1) == word2.charAt(j-1)){
                    dp[i][j] = dp[i-1][j-1];
                }
                else{
                    dp[i][j] = 1+Math.min(dp[i-1][j-1],Math.min(dp[i-1][j],dp[i][j-1])); //insert & replace: dp[i-1][j-1] +1; delete: dp[i-1][j],dp[i][j-1]
                }
            }
        }

        return dp[word1.length()][word2.length()];
    }

}