参加过的面试题目总结
1，微软实习生招聘
题目1 : String reorder 时间限制:10000ms 单点时限:1000ms 内存限制:256MB Description For this question, your program is required to process an input string containing only ASCII characters between ‘0’ and ‘9’, or between ‘a’ and ‘z’ (including ‘0’, ‘9’, ‘a’, ‘z’). Your program should reorder and split all input string characters into multiple segments, and output all segments as one concatenated string. The following requirements should also be met, 1. Characters in each segment should be in strictly increasing order. For ordering, ‘9’ is larger than ‘0’, ‘a’ is larger than ‘9’, and ‘z’ is larger than ‘a’ (basically following ASCII character order). 2. Characters in the second segment must be the same as or a subset of the first segment; and every following segment must be the same as or a subset of its previous segment. Your program should output string “<invalid input string>” when the input contains any invalid characters (i.e., outside the '0'-'9' and 'a'-'z' range). Input Input consists of multiple cases, one case per line. Each case is one string consisting of ASCII characters. Output For each case, print exactly one line with the reordered string based on the criteria above. 样例输入 aabbccdd 007799aabbccddeeff113355zz 1234.89898 abcdefabcdefabcdefaaaaaaaaaaaaaabbbbbbbddddddee 样例输出 abcdabcd 013579abcdefz013579abcdefz <invalid input string> abcdefabcdefabcdefabdeabdeabdabdabdabdabaaaaaaa
题目2 : K-th string 时间限制:10000ms 单点时限:1000ms 内存限制:256MB Description Consider a string set that each of them consists of {0, 1} only. All strings in the set have the same number of 0s and 1s. Write a program to find and output the K-th string according to the dictionary order. If such a string doesn’t exist, or the input is not valid, please output “Impossible”. For example, if we have two ‘0’s and two ‘1’s, we will have a set with 6 different strings, {0011, 0101, 0110, 1001, 1010, 1100}, and the 4th string is 1001. Input The first line of the input file contains a single integer t (1 ≤ t ≤ 10000), the number of test cases, followed by the input data for each test case. Each test case is 3 integers separated by blank space: N, M(2 <= N + M <= 33 and N , M >= 0), K(1 <= K <= 1000000000). N stands for the number of ‘0’s, M stands for the number of ‘1’s, and K stands for the K-th of string in the set that needs to be printed as output. Output For each case, print exactly one line. If the string exists, please print it, otherwise print “Impossible”. 样例输入 3 2 2 2 2 2 7 4 7 47 样例输出 0101 Impossible 01010111011 EmacsNormalVim
题目3 : Reduce inversion count 时间限制:10000ms 单点时限:1000ms 内存限制:256MB Description Find a pair in an integer array that swapping them would maximally decrease the inversion count of the array. If such a pair exists, return the new inversion count; otherwise returns the original inversion count. Definition of Inversion: Let (A[0], A[1] ... A[n], n <= 50) be a sequence of n numbers. If i < j and A[i] > A[j], then the pair (i, j) is called inversion of A. Example: Count(Inversion({3, 1, 2})) = Count({3, 1}, {3, 2}) = 2 InversionCountOfSwap({3, 1, 2})=> { InversionCount({1, 3, 2}) = 1 <-- swapping 1 with 3, decreases inversion count by 1 InversionCount({2, 1, 3}) = 1 <-- swapping 2 with 3, decreases inversion count by 1 InversionCount({3, 2, 1}) = 3 <-- swapping 1 with 2 , increases inversion count by 1 } Input Input consists of multiple cases, one case per line.Each case consists of a sequence of integers separated by comma. Output For each case, print exactly one line with the new inversion count or the original inversion count if it cannot be reduced. 样例输入 3,1,2 1,2,3,4,5 样例输出 1 0 EmacsNormalVim
题目4 : Most Frequent Logs 时间限制:10000ms 单点时限:3000ms 内存限制:256MB Description In a running system, there are many logs produced within a short period of time, we'd like to know the count of the most frequent logs. Logs are produced by a few non-empty format strings, the number of logs is N(1<=N<=20000), the maximum length of each log is 256. Here we consider a log same with another when their edit distance (see note) is <= 5. Also we have a) logs are all the same with each other produced by a certain format string b) format strings have edit distance 5 of each other. Your program will be dealing with lots of logs, so please try to keep the time cost close to O(nl), where n is the number of logs, and l is the average log length. Note edit distance is the minimum number of operations (insertdeletereplace a character) required to transform one string into the other, please refer to http://en.wikipedia.org/wiki/Edit_distance for more details. Input Multiple lines of non-empty strings. Output The count of the most frequent logs.
第一道题的解答：关键点：利用数组建立map，数组下标本身即体现key值
package com.bobo.mylianxi; import java.util.Scanner; public class Ms1 { private static String ElemStr = "0123456789abcdefghijklmnopqrstuvwxyz"; private static int char2index(char c) { return ElemStr.indexOf(c); } private static char index2char(int index) { return ElemStr.charAt(index); }; // private static int char2index(char c) { // if(c>='0'&&c<='9'){ // return c-'0'; // } // if(c>='a'&&c<='z'){ // return c-'a'+10; // } // return -1; // } // // private static char index2char(int index) { // //return ElemStr.charAt(index); // if(index>=0&&index<=9){ // return (char) ('0'+index); // } // if(index>=10&&index<=35){ // return (char) ('a'+index-10); // } // return '*'; // // }; public static void main(String[] args) { Scanner in = new Scanner(System.in); while (in.hasNext()) { String inStr = in.nextLine(); int[] timeArray = new int[36]; int flag = 1; for (int i = 0; i < inStr.length(); i++) { // 找到对应的下标并将数组中对应下标的值进行加加 int index = char2index(inStr.charAt(i)); if (index < 0) { System.out.println("invalid inputs"); // 书名此次输入的字符串不合法 flag = -1; break; } else { timeArray[index]++; } } if (flag == -1) { // 输入字符串不合法，直接判断下一个 continue; } // 之后需要将重组后的字符串进行输出 StringBuilder sb = new StringBuilder(); // 需要循环的次数为最大的那个,可以使用一个flag进行判断 boolean hasChar = true; while (hasChar) { hasChar = false; for (int i = 0; i < timeArray.length; i++) { if (timeArray[i] > 0) { sb.append(index2char(i)); timeArray[i]--; } if (timeArray[i] > 0) { hasChar = true; } } } System.out.println(sb.toString()); } } }
第二题的关键点：全排列的问题，字典序的问题
