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CS-Notes/2016校招真题题解

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Create 2016校招真题题解
2018-02-13 22:59:02 +08:00
<!-- GFM-TOC -->
* [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)
<!-- GFM-TOC -->
# 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<TreeNode> 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<Integer> 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<String> ret = new HashSet<>();
for (int i = 0; i < n; i++) {
Set<Character> 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<String, Integer> 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<Map.Entry<String, Integer>> 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<String, Integer> 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<Pair<String, Integer>> 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, K> {
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<String> paths;
private List<Integer> 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<sub>i</sub> 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]);
```
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