算法习题之DC3生成后缀数组

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));
		}

	}