介绍用DC3算法生成后缀数组的流程
1.得到S12的精确排名(取S12的前三位进行桶排序)
2.s1按照原来在数组的顺序放在左边(放第一步的排名),s2按照原来在数组的顺序放在右边中间(放第一步的排名)用最小的ASCII隔开(如果第一步得到精确的排名,跳过第2步)
3.得到s0精确排名(每一个S0的数变为S0位置的数 + 紧跟当前位置数的排名)
4.得到总排名(S0和S12进行merge,如果是S0和S1merge:S0和S1当前位置的数字做比较,下一位用S12的排名做比较,如果是S0和S2merge:S0和S1当前位置的数字做比较,下一位还用当前字符做比较,下下一位用排名做比较)
DC3模板
// arr = [a,a,b,a]
// 0 1 2 3
// 第几名在原数组的位置
// sa = [3,0,1,2]
// 0 1 2 3
// 原数组位置是第几名
// rank = [1,2,3,0]
// 0 1 2 3
public int[] sa;
public int[] rank;
public int[] height;
// 构造方法的约定:
// 数组叫nums,如果你是字符串,请转成整型数组nums
// 数组中,最小值>=1
// 如果不满足,处理成满足的,也不会影响使用
// max, nums里面最大值是多少
public DC3(int[] nums, int max) {
sa = sa(nums, max);
rank = rank();
height = height(nums);
}
private int[] sa(int[] nums, int max) {
int n = nums.length;
int[] arr = new int[n + 3];
for (int i = 0; i < n; i++) {
arr[i] = nums[i];
}
return skew(arr, n, max);
}
private int[] skew(int[] nums, int n, int K) {
int n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3, n02 = n0 + n2;
int[] s12 = new int[n02 + 3], sa12 = new int[n02 + 3];
for (int i = 0, j = 0; i < n + (n0 - n1); ++i) {
if (0 != i % 3) {
s12[j++] = i;
}
}
radixPass(nums, s12, sa12, 2, n02, K);
radixPass(nums, sa12, s12, 1, n02, K);
radixPass(nums, s12, sa12, 0, n02, K);
int name = 0, c0 = -1, c1 = -1, c2 = -1;
for (int i = 0; i < n02; ++i) {
if (c0 != nums[sa12[i]] || c1 != nums[sa12[i] + 1] || c2 != nums[sa12[i] + 2]) {
name++;
c0 = nums[sa12[i]];
c1 = nums[sa12[i] + 1];
c2 = nums[sa12[i] + 2];
}
if (1 == sa12[i] % 3) {
s12[sa12[i] / 3] = name;
} else {
s12[sa12[i] / 3 + n0] = name;
}
}
if (name < n02) {
sa12 = skew(s12, n02, name);
for (int i = 0; i < n02; i++) {
s12[sa12[i]] = i + 1;
}
} else {
for (int i = 0; i < n02; i++) {
sa12[s12[i] - 1] = i;
}
}
int[] s0 = new int[n0], sa0 = new int[n0];
for (int i = 0, j = 0; i < n02; i++) {
if (sa12[i] < n0) {
s0[j++] = 3 * sa12[i];
}
}
radixPass(nums, s0, sa0, 0, n0, K);
int[] sa = new int[n];
for (int p = 0, t = n0 - n1, k = 0; k < n; k++) {
int i = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2;
int j = sa0[p];
if (sa12[t] < n0 ? leq(nums[i], s12[sa12[t] + n0], nums[j], s12[j / 3])
: leq(nums[i], nums[i + 1], s12[sa12[t] - n0 + 1], nums[j], nums[j + 1], s12[j / 3 + n0])) {
sa[k] = i;
t++;
if (t == n02) {
for (k++; p < n0; p++, k++) {
sa[k] = sa0[p];
}
}
} else {
sa[k] = j;
p++;
if (p == n0) {
for (k++; t < n02; t++, k++) {
sa[k] = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2;
}
}
}
}
return sa;
}
private void radixPass(int[] nums, int[] input, int[] output, int offset, int n, int k) {
int[] cnt = new int[k + 1];
for (int i = 0; i < n; ++i) {
cnt[nums[input[i] + offset]]++;
}
for (int i = 0, sum = 0; i < cnt.length; ++i) {
int t = cnt[i];
cnt[i] = sum;
sum += t;
}
for (int i = 0; i < n; ++i) {
output[cnt[nums[input[i] + offset]]++] = input[i];
}
}
private boolean leq(int a1, int a2, int b1, int b2) {
return a1 < b1 || (a1 == b1 && a2 <= b2);
}
private boolean leq(int a1, int a2, int a3, int b1, int b2, int b3) {
return a1 < b1 || (a1 == b1 && leq(a2, a3, b2, b3));
}
private int[] rank() {
int n = sa.length;
int[] ans = new int[n];
for (int i = 0; i < n; i++) {
ans[sa[i]] = i;
}
return ans;
}
private int[] height(int[] s) {
int n = s.length;
int[] ans = new int[n];
for (int i = 0, k = 0; i < n; ++i) {
if (rank[i] != 0) {
if (k > 0) {
--k;
}
int j = sa[rank[i] - 1];
while (i + k < n && j + k < n && s[i + k] == s[j + k]) {
++k;
}
ans[rank[i]] = k;
}
}
return ans;
}
// 为了测试
public static int[] randomArray(int len, int maxValue) {
int[] arr = new int[len];
for (int i = 0; i < len; i++) {
arr[i] = (int) (Math.random() * maxValue) + 1;
}
return arr;
}
// 为了测试
public static void main(String[] args) {
int len = 100000;
int maxValue = 100;
long start = System.currentTimeMillis();
new DC3(randomArray(len, maxValue), maxValue);
long end = System.currentTimeMillis();
System.out.println("数据量 " + len + ", 运行时间 " + (end - start) + " ms");
}
习题1 给你一个字符串 s ,找出它的所有子串并按字典序排列,返回排在最后的那个子串
public static String lastSubstring(String s) {
if (s == null || s.length() == 0) {
return s;
}
int N = s.length();
char[] str = s.toCharArray();
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
for (char cha : str) {
min = Math.min(min, cha);
max = Math.max(max, cha);
}
int[] arr = new int[N];
for (int i = 0; i < N; i++) {
arr[i] = str[i] - min + 1;
}
DC3 dc3 = new DC3(arr, max - min + 1);
return s.substring(dc3.sa[N - 1]);
}
public static class DC3 {
public int[] sa;
public DC3(int[] nums, int max) {
sa = sa(nums, max);
}
private int[] sa(int[] nums, int max) {
int n = nums.length;
int[] arr = new int[n + 3];
for (int i = 0; i < n; i++) {
arr[i] = nums[i];
}
return skew(arr, n, max);
}
private int[] skew(int[] nums, int n, int K) {
int n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3, n02 = n0 + n2;
int[] s12 = new int[n02 + 3], sa12 = new int[n02 + 3];
for (int i = 0, j = 0; i < n + (n0 - n1); ++i) {
if (0 != i % 3) {
s12[j++] = i;
}
}
radixPass(nums, s12, sa12, 2, n02, K);
radixPass(nums, sa12, s12, 1, n02, K);
radixPass(nums, s12, sa12, 0, n02, K);
int name = 0, c0 = -1, c1 = -1, c2 = -1;
for (int i = 0; i < n02; ++i) {
if (c0 != nums[sa12[i]] || c1 != nums[sa12[i] + 1] || c2 != nums[sa12[i] + 2]) {
name++;
c0 = nums[sa12[i]];
c1 = nums[sa12[i] + 1];
c2 = nums[sa12[i] + 2];
}
if (1 == sa12[i] % 3) {
s12[sa12[i] / 3] = name;
} else {
s12[sa12[i] / 3 + n0] = name;
}
}
if (name < n02) {
sa12 = skew(s12, n02, name);
for (int i = 0; i < n02; i++) {
s12[sa12[i]] = i + 1;
}
} else {
for (int i = 0; i < n02; i++) {
sa12[s12[i] - 1] = i;
}
}
int[] s0 = new int[n0], sa0 = new int[n0];
for (int i = 0, j = 0; i < n02; i++) {
if (sa12[i] < n0) {
s0[j++] = 3 * sa12[i];
}
}
radixPass(nums, s0, sa0, 0, n0, K);
int[] sa = new int[n];
for (int p = 0, t = n0 - n1, k = 0; k < n; k++) {
int i = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2;
int j = sa0[p];
if (sa12[t] < n0 ? leq(nums[i], s12[sa12[t] + n0], nums[j], s12[j / 3])
: leq(nums[i], nums[i + 1], s12[sa12[t] - n0 + 1], nums[j], nums[j + 1], s12[j / 3 + n0])) {
sa[k] = i;
t++;
if (t == n02) {
for (k++; p < n0; p++, k++) {
sa[k] = sa0[p];
}
}
} else {
sa[k] = j;
p++;
if (p == n0) {
for (k++; t < n02; t++, k++) {
sa[k] = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2;
}
}
}
}
return sa;
}
private void radixPass(int[] nums, int[] input, int[] output, int offset, int n, int k) {
int[] cnt = new int[k + 1];
for (int i = 0; i < n; ++i) {
cnt[nums[input[i] + offset]]++;
}
for (int i = 0, sum = 0; i < cnt.length; ++i) {
int t = cnt[i];
cnt[i] = sum;
sum += t;
}
for (int i = 0; i < n; ++i) {
output[cnt[nums[input[i] + offset]]++] = input[i];
}
}
private boolean leq(int a1, int a2, int b1, int b2) {
return a1 < b1 || (a1 == b1 && a2 <= b2);
}
private boolean leq(int a1, int a2, int a3, int b1, int b2, int b3) {
return a1 < b1 || (a1 == b1 && leq(a2, a3, b2, b3));
}
}