【HackerRank】Service Lane

Calvin can enter into and exit from any segment. Let‘s call the entry segment as index i and the exit segment as index j. Assume that the exit segment lies after the entry segment(j>i) and i ≥ 0. Calvin has to pass through all segments from index i to indexj (both inclusive).

Calvin has three types of vehicles - bike, car and truck, represented by 1, 2 and 3 respectively. These numbers also denote the width of the vehicle. We are given an array width[] of length N, where width[k] represents the width of k^th segment of our service lane. It is guaranteed that while servicing he can pass through at most 1000 segments, including entry and exit segments.

• If width[k] is 1, only the bike can pass through k^th segment.

• If width[k] is 2, the bike and car can pass through k^th segment.

• If width[k] is 3, any of the bike, car or truck can pass through k^th segment.

Given the entry and exit point of Calvin‘s vehicle in the service lane, output the type of largest vehicle which can pass through the service lane (including the entry & exit segment)

Input Format
The first line of input contains two integers - N & T, where N is the length of the freeway, and T is the number of test cases. The next line has N space separated integers which represents the width array.

T test cases follow. Each test case contains two integers - i & j, where i is the index of segment through which Calvin enters the service lane and j is the index of the lane segment where he exits.

Output Format
For each test case, print the number that represents the largest vehicle type that can pass through the service lane.

Note
Calvin has to pass through all segments from index i to indexj (both inclusive).

Constraints
2 <= N <= 100000
1 <= T <= 1000
0 <= i < j < N
2 <= j-i+1 <= min(N,1000)
1 <= width[k] <= 3, where 0 <= k < N

题解：

 1 import java.io.*;
 2 import java.util.*;
 3 import java.text.*;
 4 import java.math.*;
 5 import java.util.regex.*;
 6 
 7 public class Solution {
 8    static int Service_Lane(int[] lane,int enter,int exit){
 9        int minimal = Integer.MAX_VALUE;
10        for(int i = enter;i<=exit;i++)
11            minimal = Math.min(minimal, lane[i]);
12        return minimal;
13    }
14 
15  public static void main(String[] args) {
16         Scanner in = new Scanner(System.in);
17         int n;
18         n = in.nextInt();
19         int t;
20         t = in.nextInt();
21         int[] lane = new int[n];
22         for(int i = 0;i < n;i++){
23             lane[i] = in.nextInt(); 
24         }
25         for(int i = 0;i < t;i++){
26             int enter = in.nextInt();
27             int exit = in.nextInt();
28             int mini = Service_Lane(lane, enter, exit);
29             if(mini >= 3)
30                 System.out.println(3);
31             else if(mini >= 2)
32                 System.out.println(2);
33             else
34                 System.out.println(1);
35         }
36    }
37 }

【HackerRank】Service Lane,布布扣,bubuko.com

【HackerRank】Service Lane

原文：http://www.cnblogs.com/sunshineatnoon/p/3874862.html