給一個環和一個key，判斷環經多少步可以得到key，

Input: ring = "godding", key = "gd"
Output: 4
Explanation:
 For the first key character 'g', since it is already in place, we just need 1 step to spell this character. 
 For the second key character 'd', we need to rotate the ring "godding" anticlockwise by two steps to make it become "ddinggo".
 Also, we need 1 more step for spelling.
 So the final output is 4.

public class Solution {
    public int findRotateSteps(String ring, String key) {
        int n = ring.length(), m = key.length();
        int[][] dp = new int[m + 1][n];
        int[][] clock = preproc(ring, 1), anti = preproc(ring, -1);
        for (int i = m - 1; i >= 0; --i) {
            int idx = key.charAt(i) - 'a';
            for (int j = 0; j < n; ++j) { // fill dp[i][j]
                int p = clock[j][idx];
                int q = anti[j][idx];
                // 順時針從i ~ p走過的路徑是i與p之間包含12點鐘位置的那一段！！實際轉一下圓環就知道
                dp[i][j] = Math.min(dp[i + 1][p] + (j + n - p) % n, dp[i + 1][q] + (q + n - j) % n);
            }
        }
        return dp[0][0] + m;
    }
    // ans[i]表示以i位置為視角，正/反向離char[i]最近的各字母index是多少
    int[][] preproc(String r, int inc) {
        int n = r.length();
        int[][] ans = new int[n][26];
        int[] map = new int[26];
        // 要遍歷兩個長度才能保證前面的index知道後面的值是什麼
        for (int i = 0, j = 0; j < n * 2 - 1; ++j) {
            map[r.charAt(i) - 'a'] = i;
            System.arraycopy(map, 0, ans[i], 0, 26);
            i = (i + inc + n) % n;
        }
        return ans;
    }
}