* [前言](#前言) * [1. 小米-小米Git](#1-小米-小米git) * [2. 小米-懂二进制](#2-小米-懂二进制) * [3. 小米-中国牛市](#3-小米-中国牛市) * [4. 微软-LUCKY STRING](#4-微软-lucky-string) * [5. 微软-Numeric Keypad](#5-微软-numeric-keypad) * [6. 微软-Spring Outing](#6-微软-spring-outing) * [7. 微软-S-expression](#7-微软-s-expression) * [8. 华为-最高分是多少](#8-华为-最高分是多少) * [9. 华为-简单错误记录](#9-华为-简单错误记录) * [10. 华为-扑克牌大小](#10-华为-扑克牌大小) * [11. 去哪儿-二分查找](#11-去哪儿-二分查找) * [12. 去哪儿-首个重复字符](#12-去哪儿-首个重复字符) * [13. 去哪儿-寻找Coder](#13-去哪儿-寻找coder) * [14. 美团-最大差值](#14-美团-最大差值) * [15. 美团-棋子翻转](#15-美团-棋子翻转) * [16. 美团-拜访](#16-美团-拜访) * [17. 美团-直方图内最大矩形](#17-美团-直方图内最大矩形) * [18. 美团-字符串计数](#18-美团-字符串计数) * [19. 美团-平均年龄](#19-美团-平均年龄) * [20. 百度-罪犯转移](#20-百度-罪犯转移) * [22. 百度-裁减网格纸](#22-百度-裁减网格纸) * [23. 百度-钓鱼比赛](#23-百度-钓鱼比赛) * [24. 百度-蘑菇阵](#24-百度-蘑菇阵) # 前言 省略的代码： ```java import java.util.*; ``` ```java public class Solution { } ``` ```java public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); while (in.hasNext()) { } } } ``` # 1. 小米-小米Git - 重建多叉树 - 使用 LCA ```java private class TreeNode { int id; List childs = new ArrayList<>(); TreeNode(int id) { this.id = id; } } public int getSplitNode(String[] matrix, int indexA, int indexB) { int n = matrix.length; boolean[][] linked = new boolean[n][n]; // 重建邻接矩阵 for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { linked[i][j] = matrix[i].charAt(j) == '1'; } } TreeNode tree = constructTree(linked, 0); TreeNode ancestor = LCA(tree, new TreeNode(indexA), new TreeNode(indexB)); return ancestor.id; } private TreeNode constructTree(boolean[][] linked, int root) { TreeNode tree = new TreeNode(root); for (int i = 0; i < linked[root].length; i++) { if (linked[root][i]) { linked[i][root] = false; // 因为题目给的邻接矩阵是双向的，在这里需要把它转为单向的 tree.childs.add(constructTree(links, i)); } } return tree; } private TreeNode LCA(TreeNode root, TreeNode p, TreeNode q) { if (root == null || root.id == p.id || root.id == q.id) return root; TreeNode ancestor = null; int cnt = 0; for (int i = 0; i < root.childs.size(); i++) { TreeNode tmp = LCA(root.childs.get(i), p, q); if (tmp != null) { ancestor = tmp; cnt++; } } return cnt == 2 ? root : ancestor; } ``` # 2. 小米-懂二进制 对两个数进行异或，结果的二进制表示为 1 的那一位就是两个数不同的位。 ```java public int countBitDiff(int m, int n) { return Integer.bitCount(m ^ n); } ``` # 3. 小米-中国牛市 背包问题，可以设一个大小为 2 的背包。 状态转移方程如下： ```html dp[i, j] = max(dp[i, j-1], prices[j] - prices[jj] + dp[i-1, jj]) { jj in range of [0, j-1] } = max(dp[i, j-1], prices[j] + max(dp[i-1, jj] - prices[jj])) ``` ```java public int calculateMax(int[] prices) { int n = prices.length; int[][] dp = new int[3][n]; for (int i = 1; i <= 2; i++) { int localMax = dp[i - 1][0] - prices[0]; for (int j = 1; j < n; j++) { dp[i][j] = Math.max(dp[i][j - 1], prices[j] + localMax); localMax = Math.max(localMax, dp[i - 1][j] - prices[j]); } } return dp[2][n - 1]; } ``` # 4. 微软-LUCKY STRING - 斐波那契数列可以预计算； - 从头到尾遍历字符串的过程，每一轮循环都使用一个 Set 来保存从 i 到 j 出现的字符，并且 Set 保证了字符都不同，因此 Set 的大小就是不同字符的个数。 ```java Set fibSet = new HashSet<>(Arrays.asList(1, 2, 3, 5, 8, 13, 21, 34, 55, 89)); Scanner in = new Scanner(System.in); String str = in.nextLine(); int n = str.length(); Set ret = new HashSet<>(); for (int i = 0; i < n; i++) { Set set = new HashSet<>(); for (int j = i; j < n; j++) { set.add(str.charAt(j)); int cnt = set.size(); if (fibSet.contains(cnt)) { ret.add(str.substring(i, j + 1)); } } } String[] arr = ret.toArray(new String[ret.size()]); Arrays.sort(arr); for (String s : arr) { System.out.println(s); } ``` # 5. 微软-Numeric Keypad ```java private static int[][] canReach = { {1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, // 0 {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, // 1 {1, 0, 1, 1, 0, 1, 1, 0, 1, 1}, // 2 {0, 0, 0, 1, 0, 0, 1, 0, 0, 1}, // 3 {1, 0, 0, 0, 1, 1, 1, 1, 1, 1}, // 4 {1, 0, 0, 0, 0, 1, 1, 0, 1, 1}, // 5 {0, 0, 0, 0, 0, 0, 1, 0, 0, 1}, // 6 {1, 0, 0, 0, 0, 0, 0, 1, 1, 1}, // 7 {1, 0, 0, 0, 0, 0, 0, 0, 1, 1}, // 8 {0, 0, 0, 0, 0, 0, 0, 0, 0, 1} // 9 }; private static boolean isLegal(char[] chars, int idx) { if (idx >= chars.length || idx < 0) return true; int cur = chars[idx] - '0'; int next = chars[idx + 1] - '0'; return canReach[cur][next] == 1; } public static void main(String[] args) { Scanner in = new Scanner(System.in); int T = Integer.valueOf(in.nextLine()); for (int i = 0; i < T; i++) { String line = in.nextLine(); char[] chars = line.toCharArray(); for (int j = 0; j < chars.length - 1; j++) { while (!isLegal(chars, j)) { if (--chars[j + 1] < '0') { chars[j--]--; } for (int k = j + 2; k < chars.length; k++) { chars[k] = '9'; } } } System.out.println(new String(chars)); } } ``` # 6. 微软-Spring Outing 下面以 N = 3，K = 4 来进行讨论。 初始时，令第 0 个地方成为待定地点，也就是呆在家里。 从第 4 个地点开始投票，每个人只需要比较第 4 个地方和第 0 个地方的优先级，里，如果超过半数的人选择了第 4 个地方，那么更新第 4 个地方成为待定地点。 从后往前不断重复以上步骤，不断更新待定地点，直到所有地方都已经投票。 上面的讨论中，先令第 0 个地点成为待定地点，是因为这样的话第 4 个地点就只需要和这个地点进行比较，而不用考虑其它情况。如果最开始先令第 1 个地点成为待定地点，那么在对第 2 个地点进行投票时，每个人不仅要考虑第 2 个地点与第 1 个地点的优先级，也要考虑与其后投票地点的优先级。 ```java int N = in.nextInt(); int K = in.nextInt(); int[][] votes = new int[N][K + 1]; for (int i = 0; i < N; i++) { for (int j = 0; j < K + 1; j++) { int place = in.nextInt(); votes[i][place] = j; } } int ret = 0; for (int place = K; place > 0; place--) { int cnt = 0; for (int i = 0; i < N; i++) { if (votes[i][place] < votes[i][ret]) { cnt++; } } if (cnt > N / 2) { ret = place; } } System.out.println(ret == 0 ? "otaku" : ret); ``` # 7. 微软-S-expression # 8. 华为-最高分是多少 ```java int N = in.nextInt(); int M = in.nextInt(); int[] scores = new int[N]; for (int i = 0; i < N; i++) { scores[i] = in.nextInt(); } for (int i = 0; i < M; i++) { String str = in.next(); if (str.equals("U")) { int id = in.nextInt() - 1; int newScore = in.nextInt(); scores[id] = newScore; } else { int idBegin = in.nextInt() - 1; int idEnd = in.nextInt() - 1; int ret = 0; if (idBegin > idEnd) { int t = idBegin; idBegin = idEnd; idEnd = t; } for (int j = idBegin; j <= idEnd; j++) { ret = Math.max(ret, scores[j]); } System.out.println(ret); } } ``` # 9. 华为-简单错误记录 ```java HashMap map = new LinkedHashMap<>(); while (in.hasNextLine()) { String s = in.nextLine(); String key = s.substring(s.lastIndexOf('\\') + 1); map.put(key, map.containsKey(key) ? map.get(key) + 1 : 1); } List> list = new LinkedList<>(map.entrySet()); Collections.sort(list, (o1, o2) -> o2.getValue() - o1.getValue()); for (int i = 0; i < 8 && i < list.size(); i++) { String[] token = list.get(i).getKey().split(" "); String filename = token[0]; String line = token[1]; if (filename.length() > 16) filename = filename.substring(filename.length() - 16); System.out.println(filename + " " + line + " " + list.get(i).getValue()); } ``` # 10. 华为-扑克牌大小 ```java public class Main { private Map map = new HashMap<>(); public Main() { map.put("3", 0); map.put("4", 1); map.put("5", 2); map.put("6", 3); map.put("7", 4); map.put("8", 5); map.put("9", 6); map.put("10", 7); map.put("J", 8); map.put("Q", 9); map.put("K", 10); map.put("A", 11); map.put("2", 12); map.put("joker", 13); map.put("JOKER ", 14); } private String play(String s1, String s2) { String[] token1 = s1.split(" "); String[] token2 = s2.split(" "); CardType type1 = computeCardType(token1); CardType type2 = computeCardType(token2); if (type1 == CardType.DoubleJoker) return s1; if (type2 == CardType.DoubleJoker) return s2; if (type1 == CardType.Bomb && type2 != CardType.Bomb) return s1; if (type2 == CardType.Bomb && type1 != CardType.Bomb) return s2; if (type1 != type2 || token1.length != token2.length) return "ERROR"; for (int i = 0; i < token1.length; i++) { int val1 = map.get(token1[i]); int val2 = map.get(token2[i]); if (val1 != val2) return val1 > val2 ? s1 : s2; } return "ERROR"; } private CardType computeCardType(String[] token) { boolean hasjoker = false, hasJOKER = false; for (int i = 0; i < token.length; i++) { if (token[i].equals("joker")) hasjoker = true; else if (token[i].equals("JOKER")) hasJOKER = true; } if (hasjoker && hasJOKER) return CardType.DoubleJoker; int maxContinueLen = 1; int curContinueLen = 1; String curValue = token[0]; for (int i = 1; i < token.length; i++) { if (token[i].equals(curValue)) curContinueLen++; else { curContinueLen = 1; curValue = token[i]; } maxContinueLen = Math.max(maxContinueLen, curContinueLen); } if (maxContinueLen == 4) return CardType.Bomb; if (maxContinueLen == 3) return CardType.Triple; if (maxContinueLen == 2) return CardType.Double; boolean isStraight = true; for (int i = 1; i < token.length; i++) { if (map.get(token[i]) - map.get(token[i - 1]) != 1) { isStraight = false; break; } } if (isStraight && token.length == 5) return CardType.Straight; return CardType.Sigal; } private enum CardType { DoubleJoker, Bomb, Sigal, Double, Triple, Straight; } public static void main(String[] args) { Main main = new Main(); Scanner in = new Scanner(System.in); while (in.hasNextLine()) { String s = in.nextLine(); String[] token = s.split("-"); System.out.println(main.play(token[0], token[1])); } } } ``` # 11. 去哪儿-二分查找 对于有重复元素的有序数组，二分查找需要注意以下要点： - if (val <= A[m]) h = m; - 因为 h 的赋值为 m 而不是 m - 1，因此 while 循环的条件也就为 l < h。（如果是 m - 1 循环条件为 l <= h） ```java public int getPos(int[] A, int n, int val) { int l = 0, h = n - 1; while (l < h) { int m = l + (h - l) / 2; if (val <= A[m]) h = m; else l = m + 1; } return A[h] == val ? h : -1; } ``` # 12. 去哪儿-首个重复字符 ```java public char findFirstRepeat(String A, int n) { boolean[] hasAppear = new boolean[256]; for (int i = 0; i < n; i++) { char c = A.charAt(i); if(hasAppear[c]) return c; hasAppear[c] = true; } return ' '; } ``` # 13. 去哪儿-寻找Coder ```java public String[] findCoder(String[] A, int n) { List> list = new ArrayList<>(); for (String s : A) { int cnt = 0; String t = s.toLowerCase(); int idx = -1; while (true) { idx = t.indexOf("coder", idx + 1); if (idx == -1) break; cnt++; } if (cnt != 0) { list.add(new Pair<>(s, cnt)); } } Collections.sort(list, (o1, o2) -> (o2.getValue() - o1.getValue())); String[] ret = new String[list.size()]; for (int i = 0; i < list.size(); i++) { ret[i] = list.get(i).getKey(); } return ret; } // 牛客网无法导入 javafx.util.Pair，这里就自己实现一下 Pair 类 private class Pair { T t; K k; Pair(T t, K k) { this.t = t; this.k = k; } T getKey() { return t; } K getValue() { return k; } } ``` # 14. 美团-最大差值 贪心策略。 ```java public int getDis(int[] A, int n) { int max = 0; int soFarMin = A[0]; for (int i = 1; i < n; i++) { if(soFarMin > A[i]) soFarMin = A[i]; else max = Math.max(max, A[i]- soFarMin); } return max; } ``` # 15. 美团-棋子翻转 ```java public int[][] flipChess(int[][] A, int[][] f) { int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}}; for (int[] ff : f) { for (int[] dd : direction) { int r = ff[0] + dd[0] - 1, c = ff[1] + dd[1] - 1; if(r < 0 || r > 3 || c < 0 || c > 3) continue; A[r][c] ^= 1; } } return A; } ``` # 16. 美团-拜访 ```java private Set paths; private List curPath; public int countPath(int[][] map, int n, int m) { paths = new HashSet<>(); curPath = new ArrayList<>(); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (map[i][j] == 1) { map[i][j] = -1; int[][] leftRightDirection = {{1, 0}, {-1, 0}}; int[][] topDownDirection = {{0, 1}, {0, -1}}; for (int[] lr : leftRightDirection) { for (int[] td : topDownDirection) { int[][] directions = {lr, td}; backtracking(map, n, m, i, j, directions); } } return paths.size(); } } } return 0; } private void backtracking(int[][] map, int n, int m, int r, int c, int[][] directions) { if (map[r][c] == 2) { String path = ""; for (int num : curPath) { path += num; } paths.add(path); return; } for (int i = 0; i < directions.length; i++) { int nextR = r + directions[i][0]; int nextC = c + directions[i][1]; if (nextR < 0 || nextR >= n || nextC < 0 || nextC >= m || map[nextR][nextC] == -1) continue; map[nextR][nextC] = map[nextR][nextC] == 2 ? 2 : -1; curPath.add(nextR); curPath.add(nextC); backtracking(map, n, m, nextR, nextC, directions); curPath.remove(curPath.size() - 1); curPath.remove(curPath.size() - 1); map[nextR][nextC] = map[nextR][nextC] == 2 ? 2 : 0; } } ``` # 17. 美团-直方图内最大矩形 ```java public int countArea(int[] A, int n) { int max = 0; for (int i = 0; i < n; i++) { int min = A[i]; for (int j = i; j < n; j++) { min = Math.min(min, A[j]); max = Math.max(max, min * (j - i + 1)); } } return max; } ``` # 18. 美团-字符串计数 字符串都是小写字符，可以把字符串当成是 26 进制。但是字典序的比较和普通的整数比较不同，是从左往右进行比较，例如 "ac" 和 "abc"，字典序的比较结果为 "ac" > "abc"，如果按照整数方法比较，因为 "abc" 是三位数，显然更大。 由于两个字符串的长度可能不想等，在 s1 空白部分和 s2 对应部分进行比较时，应该把 s1 的空白部分看成是 'a' 字符进行填充的。 还有一点要注意的是，s1 到 s2 长度为 len_i 的字符串个数只比较前面 i 个字符。例如 'aaa' 和 'bbb' ，长度为 2 的个数为 'aa' 到 'bb' 的字符串个数，不需要考虑后面部分的字符。 在统计个数时，从 len1 开始一直遍历到最大合法长度，每次循环都统计长度为 i 的子字符串个数。 ```java String s1 = in.next(); String s2 = in.next(); int len1 = in.nextInt(); int len2 = in.nextInt(); int len = Math.min(s2.length(), len2); int[] subtractArr = new int[len]; for (int i = 0; i < len; i++) { char c1 = i < s1.length() ? s1.charAt(i) : 'a'; char c2 = s2.charAt(i); subtractArr[i] = c2 - c1; } int ret = 0; for (int i = len1; i <= len; i++) { for (int j = 0; j < i; j++) { ret += subtractArr[j] * Math.pow(26, i - j - 1); } } System.out.println(ret - 1); ``` # 19. 美团-平均年龄 ```java int W = in.nextInt(); double Y = in.nextDouble(); double x = in.nextDouble(); int N = in.nextInt(); while (N-- > 0) { Y++; // 老员工每年年龄都要加 1 Y += (21 - Y) * x; } System.out.println((int) Math.ceil(Y)); ``` # 20. 百度-罪犯转移 部分和问题，将每次求的部分和缓存起来。 ```java int n = in.nextInt(); int t = in.nextInt(); int c = in.nextInt(); int[] values = new int[n]; for (int i = 0; i < n; i++) { values[i] = in.nextInt(); } int cnt = 0; int totalValue = 0; for (int s = 0, e = c - 1; e < n; s++, e++) { if (s == 0) { for (int j = 0; j < c; j++) totalValue += values[j]; } else { totalValue = totalValue - values[s - 1] + values[e]; } if (totalValue <= t) cnt++; } System.out.println(cnt); ``` # 22. 百度-裁减网格纸 ```java int n = in.nextInt(); int minX, minY, maxX, maxY; minX = minY = Integer.MAX_VALUE; maxX = maxY = Integer.MIN_VALUE; for (int i = 0; i < n; i++) { int x = in.nextInt(); int y = in.nextInt(); minX = Math.min(minX, x); minY = Math.min(minY, y); maxX = Math.max(maxX, x); maxY = Math.max(maxY, y); } System.out.println((int) Math.pow(Math.max(maxX - minX, maxY - minY), 2)); ``` # 23. 百度-钓鱼比赛 P ( 至少钓一条鱼 ) = 1 - P ( 一条也钓不到 ) 坑：读取概率矩阵的时候，需要一行一行进行读取，而不能直接用 in.nextDouble()。 ```java public static void main(String[] args) { Scanner in = new Scanner(System.in); while (in.hasNext()) { int n = in.nextInt(); int m = in.nextInt(); int x = in.nextInt(); int y = in.nextInt(); int t = in.nextInt(); in.nextLine(); // 坑 double pcc = 0.0; double sum = 0.0; for (int i = 1; i <= n; i++) { String[] token = in.nextLine().split(" "); // 坑 for (int j = 1; j <= m; j++) { double p = Double.parseDouble(token[j - 1]); // double p = in.nextDouble(); sum += p; if (i == x && j == y) { pcc = p; } } } double pss = sum / (n * m); pcc = computePOfIRT(pcc, t); pss = computePOfIRT(pss, t); System.out.println(pcc > pss ? "cc" : pss > pcc ? "ss" : "equal"); System.out.printf("%.2f\n", Math.max(pcc, pss)); } } // compute probability of independent repeated trials private static double computePOfIRT(double p, int t) { return 1 - Math.pow((1 - p), t); } ``` # 24. 百度-蘑菇阵 这题用回溯会超时，需要用 DP。 dp[i][j] 表示到达 (i,j) 位置不会触碰蘑菇的概率。对于 N\*M 矩阵，如果 i == N || j == M，那么 (i,j) 只能有一个移动方向；其它情况下能有两个移动方向。 考虑以下矩阵，其中第 3 行和第 3 列只能往一个方向移动，而其它位置可以有两个方向移动。 ```java int N = in.nextInt(); int M = in.nextInt(); int K = in.nextInt(); boolean[][] mushroom = new boolean[N][M]; while (K-- > 0) { int x = in.nextInt(); int y = in.nextInt(); mushroom[x - 1][y - 1] = true; } double[][] dp = new double[N][M]; dp[0][0] = 1; for (int i = 0; i < N; i++) { for (int j = 0; j < M; j++) { if (mushroom[i][j]) dp[i][j] = 0; else { double cur = dp[i][j]; if (i == N - 1 && j == M - 1) break; if (i == N - 1) dp[i][j + 1] += cur; else if (j == M - 1) dp[i + 1][j] += cur; else { dp[i][j + 1] += cur / 2; dp[i + 1][j] += cur / 2; } } } } System.out.printf("%.2f\n", dp[N - 1][M - 1]); ```