更改字符串以使它们相等

Question

参考这里的问题

我们有两个字符串A和B,它们具有相同的超级字符集.我们需要更改这些字符串以获得两个相等的字符串.在每次移动中,我们都可以执行以下操作之一:

1-交换字符串的两个连续字符
2-交换字符串的第一个和最后一个字符

可以对任一字符串执行移动.为获得两个相等的字符串,我们需要的最小移动次数是多少？输入格式和约束:输入的第一行和第二行包含两个字符串A和B.保证它们的字符相等的超集.1 <= length(A)= length(B)<= 2000所有输入字符都在'a'和'z'之间

看起来这必须使用动态编程来解决.但我无法想出方程式.有人建议他们回答 - 但看起来并不合适.

dp[i][j] = 
Min{ 
dp[i + 1][j - 1] + 1, if str1[i] = str2[j] && str1[j] = str2[i]
dp[i + 2][j] + 1, if str1[i] = str2[i + 1] && str1[i + 1] = str2[i]
dp[i][j - 2] + 1, if str1[j] = str2[j - 1] && str1[j - 1] = str2[j]
}
In short, it's 
dp[i][j] = Min(dp[i + 1][j - 1], dp[i + 2][j], dp[i][j - 2]) + 1.
Here dp[i][j] means the number of minimum swaps needs to swap str1[i, j] to str2[i, j]. Here str1[i, j] means the substring of str1 starting from pos i to pos j :)

Here is an example like the one in the quesition,
str1 = "aab",
str2 = "baa"

dp[1][1] = 0 since str1[1] == str2[1];
dp[0][2] = str1[0 + 1][2 - 1] + 1 since str1[0] = str2[2] && str1[2] = str2[0].

Answer 1

您有两个原子操作：

1. 连续交换成本为 1

2. 以成本 1 交换第一个和最后一个

一个有趣的事实：

1. 和 2. 如果字符串 end 将连接到字符串 begin （循环字符串），则它们是相同的

所以我们可以推导出更通用的操作

3. 以 cost = |from - to| 移动角色 （跨越国界）

对我来说，这个问题似乎不是二维的，或者说我无法确定维度。将此算法视为简单的方法：

private static int transform(String from, String to) {
    int commonLength = to.length();
    List<Solution> worklist = new ArrayList<>();
    worklist.add(new Solution(0,from));
    while (!worklist.isEmpty()) {
        Solution item = worklist.remove(0);
        if (item.remainder.length() == 0) {
            return item.cost;
        } else {
            int toPosition = commonLength - item.remainder.length();
            char expected = to.charAt(toPosition);
            nextpos : for (int i = 0; i < item.remainder.length(); i++) {
                if (item.remainder.charAt(i) == expected) {
                    Solution nextSolution = item.moveCharToBegin(i, commonLength);
                    for (Solution solution : worklist) {
                        if (solution.remainder.equals(nextSolution.remainder)) {
                            solution.cost = Math.min(solution.cost, nextSolution.cost);
                            continue nextpos;
                        }
                    }
                    worklist.add(nextSolution);
                }
            }
        }
    }
    return Integer.MAX_VALUE;
}

private static class Solution {
    public int cost;
    public String remainder;

    public Solution(int cost, String remainder) {
        this.cost = cost;
        this.remainder = remainder;
    }

    public Solution moveCharToBegin(int i, int length) {
        int costOffset = Math.min(i, length - i); //minimum of forward and backward circular move
        String newRemainder = remainder.substring(0, i) + remainder.substring(i + 1);
        return new Solution(cost + costOffset, newRemainder);
    }
}