72. Edit Distance

    /*
     * 72. Edit Distance 
     * 12.10 by Mingyang 
     * dp[i][j]表示从word1前i个字符转换到word2前j个字符最少的步骤数。
     * 假设word1现在遍历到字符x，word2遍历到字符y（word1当前遍历到的长度为i，word2为j）。 以下两种可能性： 1.
     * x==y，那么不用做任何编辑操作，所以dp[i][j] = dp[i-1][j-1] 
     * 2. x != y 
     * (1) 在word1插入y，那么dp[i][j] = dp[i][j-1] + 1 
     * (2) 在word1删除x， 那么dp[i][j] = dp[i-1][j] + 1
     * (3) 把word1中的x用y来替换，那么dp[i][j] = dp[i-1][j-1] + 1 最少的步骤就是取这三个中的最小值。
     * 一点关于DP的小总结，一定要设定len+1个数，因为dp[0][0]没什么意义，最后还是求的dp[len1][len2]
     * 然后记住，一个string为0的时候另一个就全是i
     */    
    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 {
                    int replace = dp[i - 1][j - 1] + 1;
                    int insert = dp[i - 1][j] + 1;
                    int delete = dp[i][j - 1] + 1;
                    int min = Math.min(replace, insert);
                    min = Math.min(min, delete);
                    dp[i][j] = min;
                }
            }
        }
        return dp[len1][len2];
    }