思路: 模拟

$O(n^2)$全遍历即可

class Solution {public int countPrefixSuffixPairs(String[] words) {int res = 0;for (int i = 0; i < words.length; i++) {for (int j = 0; j < i; j++) {if (words[i].startsWith(words[j]) && words[i].endsWith(words[j])) {res ++;}}}return res;}
}

思路: 字典树

这样的时间复杂度为

$$(ma{x}_{ai}len(wor{d}_{ai}))\ast (\sum _{bi}len(wor{d}_{bi}))$$

因为第一组的数字最大为8位数，因此其深度最多为8

public class Solution {static class Trie {Trie[] cs = new Trie[26];void add(String w) {Trie cur = this;for (char c: w.toCharArray()) {int p = c - '0';if (cur.cs[p] == null) {cur.cs[p] = new Trie();}cur = cur.cs[p];}}int query(String w) {Trie cur = this;int res = 0;for (char c: w.toCharArray()) {int p = c - '0';if (cur.cs[p] == null) {return res;}cur = cur.cs[p];res++;}return res;}}public int longestCommonPrefix(int[] arr1, int[] arr2) {Trie root = new Trie();for (int v: arr1) {root.add(String.valueOf(v));}int res = 0;for (int v: arr2) {int tmp = root.query(String.valueOf(v));res = Math.max(tmp, res);}return res;}}

思路: 模拟

知识点: 质数筛，质数判定

按题目模拟执行即可

class Solution {static int[][] dirs = new int[][] {{-1, 0}, {1, 0}, {0, -1}, {0, 1},{-1, -1}, {-1, 1}, {1, -1}, {1, 1}};List<Integer> create(int[][] mat, int y, int x) {List<Integer> res = new ArrayList<>();int n = mat.length, m = mat[0].length;for (int i = 0; i < dirs.length; i++) {int v = mat[y][x];int j = 0;int ty = y, tx = x;int base = 10;do {ty += dirs[i][0];tx += dirs[i][1];if (ty >= 0 && ty < n && tx >= 0 && tx < m) {v = v + mat[ty][tx] * base;} else {break;}base *= 10;res.add(v);} while(true);}return res;}boolean isPrime(int v) {if (v <= 1) return false;for (int i = 2; i <= v / i; i++) {if (v % i == 0) return false;}return true;}public int mostFrequentPrime(int[][] mat) {Map<Integer, Integer> hash = new HashMap<>();int n = mat.length, m = mat[0].length;for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {List<Integer> tmp = create(mat, i, j);for (int v: tmp) {hash.merge(v, 1, Integer::sum);}}}    int fz = -1;int ans = -1;for (var kv: hash.entrySet()) {if (isPrime(kv.getKey()) && kv.getKey() > 10) {if (fz < kv.getValue() || (fz == kv.getValue() && ans < kv.getKey())) {fz = kv.getValue();ans = kv.getKey();}}}return ans;}}

思路: 字典树+z函数

因为涉及到前后缀一致的问题

所以必然需要引入

• z函数
• stringhash

这类技巧

有涉及到前缀问题，自然就引出字典树

class Solution {static int[] zFunction(String a) {char[] s = a.toCharArray();int n = s.length;int[] z = new int[n];z[0] = n;for (int i = 1, l = 0, r = 0; i < n; ++i) {if (i <= r && z[i - l] < r - i + 1) {z[i] = z[i - l];} else {z[i] = Math.max(0, r - i + 1);while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];}if (i + z[i] - 1 > r) {l = i;r = i + z[i] - 1;}}return z;}static class Trie {Trie[] cs = new Trie[26];int rel;void add(String w) {Trie cur = this;for (char c: w.toCharArray()) {int p = c - 'a';if (cur.cs[p] == null) {cur.cs[p] = new Trie();}cur = cur.cs[p];}cur.rel++;}int count(String w) {Trie cur = this;int[] zz = zFunction(w);int res = 0;int n = w.length();for (int i = 0; i < n; i++) {int p = w.charAt(i) - 'a';if (cur.cs[p] == null) {return res;}cur = cur.cs[p];if (zz[n - 1 - i] == i + 1) {res += cur.rel;}}return res;}}public long countPrefixSuffixPairs(String[] words) {long res = 0;Trie root = new Trie();for (String w: words) {res += root.count(w);root.add(w);}return res;}}