力扣第 383 场周赛解题报告 | 珂学家

本文主要是介绍力扣第 383 场周赛解题报告 | 珂学家 | Z函数/StringHash，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

前言

谁言别后终无悔
寒月清宵绮梦回
深知身在情长在
前尘不共彩云飞

整体评价

T3是道模拟题，但是感觉题意有些晦涩，T4一眼Z函数，当然StringHash更通用些。

新年快乐, ^_.

T1. 将单词恢复初始状态所需的最短时间 I

思路: 模拟

就是前缀和为0的次数

class Solution {public int returnToBoundaryCount(int[] nums) {int acc = 0;int res = 0;for (int v: nums) {acc += v;if (acc == 0) res++;}return res;}
}

T2. 将单词恢复初始状态所需的最短时间 I

和T4一起讲

T3. 找出网格的区域平均强度

思路: 模拟

感觉题意有些晦涩，窗口比较小，可以莽些。

如果窗口大小比较大，就需要

$二维前缀和 + 二维差分$

class Solution {int[][] dirs = new int[][] {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};int[] evaluate(int[][] image, int ty, int tx, int limit) {int res = 0;for (int i = 0; i < 3; i++) {for (int j = 0; j < 3; j++) {for (int k = 0; k < 4; k++) {int dy = i + dirs[k][0];int dx = j + dirs[k][1];if (dy >= 0 && dy < 3 && dx >= 0 && dx < 3) {int dist = Math.abs(image[ty+i][tx+j] - image[ty+dy][tx+dx]);if (dist > limit) {return new int[] {0, 0};}}}res += image[ty + i][tx + j];}}return new int[] {1, res / 9};}public int[][] resultGrid(int[][] image, int threshold) {int h = image.length, w = image[0].length;int[][] sum = new int[h][w];int[][] cnt = new int[h][w];// 预处理，枚举左上角for (int i = 0; i < h - 2; i++) {for (int j = 0; j < w - 2; j++) {int[] res = evaluate(image, i, j, threshold);for (int s = 0; s < 3; s++) {for (int t = 0; t < 3; t++) {sum[s+i][t+j] += res[1];cnt[s+i][t+j] += res[0];}}}}// 统计值int[][] ans = new int[h][w];for (int i = 0; i < h; i++) {for (int j = 0; j < w; j++) {if (cnt[i][j] == 0) {ans[i][j] = image[i][j];} else {ans[i][j] = sum[i][j] / cnt[i][j];}}}return ans;}
}

T4. 将单词恢复初始状态所需的最短时间 II

思路: Z函数/StringHash

属于思维题的范畴

$S [i : n] == S [0 : n - i] ，那么该 i 就是满足要求的点$

因为字符串S很大，所以利用StringHash，可以 $O (N)$ 判定。

当然也可以转换下思维

$S [i : n] == S [0 : n - i] 本质就是求 S [i : n] 和 S [0 : n] 的最长前缀长度$

这个就是属于Z函数的核心概念了。

Z函数

class Solution {// z函数模板, https://oi-wiki.org/string/z-func/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;}public int minimumTimeToInitialState(String word, int k) {int n = word.length();int[] arr = zFunction(word);for (int i = k; i < n; i += k) {if (arr[i] == n - i) {return i / k;}}return (n + k - 1) / k;}}

StringHash

class Solution {staticclass StringHash {char[] str;long p, mod;int n;long[] pre; // hash前缀和long[] pow; // p的幂次public StringHash(String s, int p, long mod) {this.str = s.toCharArray();this.p = p;this.mod = mod;this.n = str.length;pre = new long[n + 1];pow = new long[n + 1];pow[0] = 1;for (int i = 1; i <= n; i++) {pow[i] = pow[i - 1] * p % mod;}for (int i = 0; i < str.length; i++) {pre[i + 1] = (pre[i] * p % mod + str[i]) % mod;}}long query(int l, int r) {long res = pre[r + 1] - pre[l] * pow[r - l + 1] % mod;return (res % mod + mod) % mod;}long rotate(int l) {if (l < 0 || l >= str.length - 1) {return query(0, str.length - 1);} else {long h1 = query(0, l);long h2 = query(l + 1, str.length - 1);return (h2 * pow[l + 1] % mod + h1) % mod;}}static long evaluate(String s, int p, long mod) {long h = 0;for (char c: s.toCharArray()) {h = (h * p % mod + c) % mod;}return h;}}public int minimumTimeToInitialState(String word, int k) {int p1 = 13, p2 = 17;long mod = (long)1e9 + 7;StringHash h1 = new StringHash(word, p1, mod);StringHash h2 = new StringHash(word, p2, mod);int res = 1;int n = word.length();for (int i = k; i < n; i += k) {int d = n - i;if (h1.query(i, n - 1) == h1.query(0, d - 1)&& h2.query(i, n - 1) == h2.query(0, d - 1)) {return res;}res++;}return res;}
}