diff --git a/2016校招真题题解 b/2016校招真题题解 new file mode 100644 index 00000000..ad4ddf29 --- /dev/null +++ b/2016校招真题题解 @@ -0,0 +1,741 @@ + +* [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 leni 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]); +```