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package class36;
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// 来自网易
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// 规定：L[1]对应a，L[2]对应b，L[3]对应c，...，L[25]对应y
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// S1 = a
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// S(i) = S(i-1) + L[i] + reverse(invert(S(i-1)));
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// 解释invert操作：
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// S1 = a
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// S2 = aby
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// 假设invert(S(2)) = 甲乙丙
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// a + 甲 = 26, 那么 甲 = 26 - 1 = 25 -> y
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// b + 乙 = 26, 那么 乙 = 26 - 2 = 24 -> x
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// y + 丙 = 26, 那么 丙 = 26 - 25 = 1 -> a
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// 如上就是每一位的计算方式，所以invert(S2) = yxa
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// 所以S3 = S2 + L[3] + reverse(invert(S2)) = aby + c + axy = abycaxy
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// invert(abycaxy) = yxawyba, 再reverse = abywaxy
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// 所以S4 = abycaxy + d + abywaxy = abycaxydabywaxy
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// 直到S25结束
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// 给定两个参数n和k，返回Sn的第k位是什么字符，n从1开始，k从1开始
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// 比如n=4，k=2，表示S4的第2个字符是什么，返回b字符
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public class Code01_ReverseInvertString {
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public static int[] lens = null;
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public static void fillLens() {
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lens = new int[26];
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lens[1] = 1;
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for (int i = 2; i <= 25; i++) {
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lens[i] = (lens[i - 1] << 1) + 1;
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}
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}
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// 求sn中的第k个字符
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// O(n), s <= 25 O(1)
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public static char kth(int n, int k) {
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if (lens == null) {
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fillLens();
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}
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if (n == 1) { // 无视k
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return 'a';
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}
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// sn half
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int half = lens[n - 1];
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if (k <= half) {
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return kth(n - 1, k);
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} else if (k == half + 1) {
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return (char) ('a' + n - 1);
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} else {
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// sn
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// 我需要右半区，从左往右的第a个
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// 需要找到，s(n-1)从右往左的第a个
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// 当拿到字符之后，invert一下，就可以返回了！
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return invert(kth(n - 1, ((half + 1) << 1) - k));
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}
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}
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public static char invert(char c) {
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return (char) (('a' << 1) + 24 - c);
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}
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// 为了测试
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public static String generateString(int n) {
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String s = "a";
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for (int i = 2; i <= n; i++) {
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s = s + (char) ('a' + i - 1) + reverseInvert(s);
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}
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return s;
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}
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// 为了测试
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public static String reverseInvert(String s) {
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char[] invert = invert(s).toCharArray();
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for (int l = 0, r = invert.length - 1; l < r; l++, r--) {
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char tmp = invert[l];
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invert[l] = invert[r];
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invert[r] = tmp;
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}
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return String.valueOf(invert);
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}
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// 为了测试
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public static String invert(String s) {
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char[] str = s.toCharArray();
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for (int i = 0; i < str.length; i++) {
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str[i] = invert(str[i]);
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}
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return String.valueOf(str);
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}
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// 为了测试
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public static void main(String[] args) {
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int n = 20;
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String str = generateString(n);
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int len = str.length();
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System.out.println("测试开始");
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for (int i = 1; i <= len; i++) {
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if (str.charAt(i - 1) != kth(n, i)) {
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System.out.println(i);
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System.out.println(str.charAt(i - 1));
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System.out.println(kth(n, i));
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System.out.println("出错了！");
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break;
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}
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}
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System.out.println("测试结束");
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}
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}
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