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137 lines
3.4 KiB
137 lines
3.4 KiB
package class44;
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// 测试链接: https://leetcode.com/problems/last-substring-in-lexicographical-order/
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public class Code01_LastSubstringInLexicographicalOrder {
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public static String lastSubstring(String s) {
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if (s == null || s.length() == 0) {
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return s;
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}
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int N = s.length();
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char[] str = s.toCharArray();
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int min = Integer.MAX_VALUE;
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int max = Integer.MIN_VALUE;
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for (char cha : str) {
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min = Math.min(min, cha);
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max = Math.max(max, cha);
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}
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int[] arr = new int[N];
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for (int i = 0; i < N; i++) {
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arr[i] = str[i] - min + 1;
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}
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DC3 dc3 = new DC3(arr, max - min + 1);
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return s.substring(dc3.sa[N - 1]);
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}
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public static class DC3 {
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public int[] sa;
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public DC3(int[] nums, int max) {
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sa = sa(nums, max);
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}
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private int[] sa(int[] nums, int max) {
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int n = nums.length;
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int[] arr = new int[n + 3];
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for (int i = 0; i < n; i++) {
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arr[i] = nums[i];
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}
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return skew(arr, n, max);
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}
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private int[] skew(int[] nums, int n, int K) {
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int n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3, n02 = n0 + n2;
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int[] s12 = new int[n02 + 3], sa12 = new int[n02 + 3];
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for (int i = 0, j = 0; i < n + (n0 - n1); ++i) {
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if (0 != i % 3) {
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s12[j++] = i;
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}
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}
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radixPass(nums, s12, sa12, 2, n02, K);
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radixPass(nums, sa12, s12, 1, n02, K);
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radixPass(nums, s12, sa12, 0, n02, K);
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int name = 0, c0 = -1, c1 = -1, c2 = -1;
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for (int i = 0; i < n02; ++i) {
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if (c0 != nums[sa12[i]] || c1 != nums[sa12[i] + 1] || c2 != nums[sa12[i] + 2]) {
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name++;
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c0 = nums[sa12[i]];
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c1 = nums[sa12[i] + 1];
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c2 = nums[sa12[i] + 2];
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}
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if (1 == sa12[i] % 3) {
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s12[sa12[i] / 3] = name;
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} else {
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s12[sa12[i] / 3 + n0] = name;
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}
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}
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if (name < n02) {
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sa12 = skew(s12, n02, name);
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for (int i = 0; i < n02; i++) {
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s12[sa12[i]] = i + 1;
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}
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} else {
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for (int i = 0; i < n02; i++) {
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sa12[s12[i] - 1] = i;
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}
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}
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int[] s0 = new int[n0], sa0 = new int[n0];
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for (int i = 0, j = 0; i < n02; i++) {
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if (sa12[i] < n0) {
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s0[j++] = 3 * sa12[i];
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}
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}
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radixPass(nums, s0, sa0, 0, n0, K);
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int[] sa = new int[n];
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for (int p = 0, t = n0 - n1, k = 0; k < n; k++) {
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int i = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2;
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int j = sa0[p];
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if (sa12[t] < n0 ? leq(nums[i], s12[sa12[t] + n0], nums[j], s12[j / 3])
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: leq(nums[i], nums[i + 1], s12[sa12[t] - n0 + 1], nums[j], nums[j + 1], s12[j / 3 + n0])) {
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sa[k] = i;
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t++;
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if (t == n02) {
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for (k++; p < n0; p++, k++) {
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sa[k] = sa0[p];
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}
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}
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} else {
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sa[k] = j;
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p++;
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if (p == n0) {
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for (k++; t < n02; t++, k++) {
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sa[k] = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2;
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}
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}
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}
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}
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return sa;
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}
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private void radixPass(int[] nums, int[] input, int[] output, int offset, int n, int k) {
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int[] cnt = new int[k + 1];
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for (int i = 0; i < n; ++i) {
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cnt[nums[input[i] + offset]]++;
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}
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for (int i = 0, sum = 0; i < cnt.length; ++i) {
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int t = cnt[i];
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cnt[i] = sum;
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sum += t;
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}
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for (int i = 0; i < n; ++i) {
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output[cnt[nums[input[i] + offset]]++] = input[i];
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}
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}
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private boolean leq(int a1, int a2, int b1, int b2) {
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return a1 < b1 || (a1 == b1 && a2 <= b2);
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}
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private boolean leq(int a1, int a2, int a3, int b1, int b2, int b3) {
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return a1 < b1 || (a1 == b1 && leq(a2, a3, b2, b3));
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}
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}
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}
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