DP动态规划专题十三：LeetCode 72. Edit Distance

LeetCode 72. Edit Distance

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:

Insert a character
Delete a character
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')

每次的操作有三种，也可能没操作，所以分析case，即可得出结论。

        public int minDistance(String word1, String word2) {
        int n = word1.length();
        int m = word2.length();
        int[][] dp = new int[n+1][m+1];
        for (int i = 1; i <= n; i++) dp[i][0] = i;
        for (int i = 1; i <= m; i++) dp[0][i] = i;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; 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(Math.min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]);
                }
            }
        }
        for (int i =0; i <= n;i++) {
            for(int j = 0; j<=m;j++) {
                System.out.print(dp[i][j] +"  ");
            }
            System.out.println();
        }
        return dp[n][m];
    }

空间优化版：

    public int minDistance(String word1, String word2) {
        int n = word1.length();
        int m = word2.length();
        int[] pre = new int[m+1];
        int[] cur = new int[m+1];
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= m; j++) {
                if (i == 0 && j == 0) cur[j] = 0;
                else if (i == 0) cur[j] = j;
                else if (j == 0) cur[j] = i;
                else{
                    if (word1.charAt(i-1) == word2.charAt(j-1)) {
                        cur[j] = pre[j-1];
                    } else {
                        cur[j] = 1 + Math.min(Math.min(cur[j-1], pre[j]), pre[j-1]);
                    }     
                }
            }
            int[] tmp = pre;
            pre = cur;
            cur = tmp;
        }
        return pre[m];
    }