package class45; // 最长公共子串问题是面试常见题目之一 // 假设str1长度N,str2长度M // 因为最优解的难度所限,一般在面试场上回答出O(N*M)的解法已经是比较优秀了 // 因为得到O(N*M)的解法,就已经需要用到动态规划了 // 但其实这个问题的最优解是O(N+M),为了达到这个复杂度可是不容易 // 首先需要用到DC3算法得到后缀数组(sa) // 进而用sa数组去生成height数组 // 而且在生成的时候,还有一个不回退的优化,都非常不容易理解 // 这就是后缀数组在面试算法中的地位 : 德高望重的噩梦 public class Code03_LongestCommonSubstringConquerByHeight { public static int lcs1(String s1, String s2) { if (s1 == null || s2 == null || s1.length() == 0 || s2.length() == 0) { return 0; } char[] str1 = s1.toCharArray(); char[] str2 = s2.toCharArray(); int row = 0; int col = str2.length - 1; int max = 0; while (row < str1.length) { int i = row; int j = col; int len = 0; while (i < str1.length && j < str2.length) { if (str1[i] != str2[j]) { len = 0; } else { len++; } if (len > max) { max = len; } i++; j++; } if (col > 0) { col--; } else { row++; } } return max; } public static int lcs2(String s1, String s2) { if (s1 == null || s2 == null || s1.length() == 0 || s2.length() == 0) { return 0; } char[] str1 = s1.toCharArray(); char[] str2 = s2.toCharArray(); int N = str1.length; int M = str2.length; int min = str1[0]; int max = str1[0]; for (int i = 1; i < N; i++) { min = Math.min(min, str1[i]); max = Math.max(max, str1[i]); } for (int i = 0; i < M; i++) { min = Math.min(min, str2[i]); max = Math.max(max, str2[i]); } int[] all = new int[N + M + 1]; int index = 0; for (int i = 0; i < N; i++) { all[index++] = str1[i] - min + 2; } all[index++] = 1; for (int i = 0; i < M; i++) { all[index++] = str2[i] - min + 2; } DC3 dc3 = new DC3(all, max - min + 2); int n = all.length; int[] sa = dc3.sa; int[] height = dc3.height; int ans = 0; for (int i = 1; i < n; i++) { int Y = sa[i - 1]; int X = sa[i]; if (Math.min(X, Y) < N && Math.max(X, Y) > N) { ans = Math.max(ans, height[i]); } } return ans; } public static class DC3 { public int[] sa; public int[] rank; public int[] height; 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]; // 依次求h[i] , k = 0 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; } // h[i] = k ans[rank[i]] = k; } } return ans; } } // for test public static String randomNumberString(int len, int range) { char[] str = new char[len]; for (int i = 0; i < len; i++) { str[i] = (char) ((int) (Math.random() * range) + 'a'); } return String.valueOf(str); } public static void main(String[] args) { int len = 30; int range = 5; int testTime = 100000; System.out.println("功能测试开始"); for (int i = 0; i < testTime; i++) { int N1 = (int) (Math.random() * len); int N2 = (int) (Math.random() * len); String str1 = randomNumberString(N1, range); String str2 = randomNumberString(N2, range); int ans1 = lcs1(str1, str2); int ans2 = lcs2(str1, str2); if (ans1 != ans2) { System.out.println("Oops!"); } } System.out.println("功能测试结束"); System.out.println("=========="); System.out.println("性能测试开始"); len = 80000; range = 26; long start; long end; String str1 = randomNumberString(len, range); String str2 = randomNumberString(len, range); start = System.currentTimeMillis(); int ans1 = lcs1(str1, str2); end = System.currentTimeMillis(); System.out.println("方法1结果 : " + ans1 + " , 运行时间 : " + (end - start) + " ms"); start = System.currentTimeMillis(); int ans2 = lcs2(str1, str2); end = System.currentTimeMillis(); System.out.println("方法2结果 : " + ans2 + " , 运行时间 : " + (end - start) + " ms"); System.out.println("性能测试结束"); } }