Groupon面经Prepare: Max Cycle Length

题目是遇到偶数/2，遇到奇数 *3 + 1的题目，然后找一个range内所有数字的max cycle length。对于一个数字，比如说44，按照题目的公式不停计算，过程是 44, 22, 11, 8, 9 ,1（瞎起的），从44到1的这个sequence的长度，叫做cycle length。然后题目是给一个range，比如[2,300]，求这里面所有数字的cycle length的最大值。follow up跑1到1 million

一个数一个数慢慢算，如下算法求出每个数的cycle length

 1 package Sorting;
 2 import java.util.*;
 3
 4 public class Solution2 {
 5     public List<Integer> maxCycleLen(int min, int max) {
 6         List<Integer> maxCycle = new ArrayList<Integer>();
 7         for (int i=min; i<=max; i++) {
 8             helper(maxCycle, i);
 9         }
10         return maxCycle;
11     }
12
13     public void helper(List<Integer> maxCycle, int num) {
14         int len = 1;
15         while (num != 1) {
16             if (num%2 == 1) num = num*3+1;
17             else num = num/2;
18             len++;
19         }
20         maxCycle.add(len);
21     }
22
23     /**
24      * @param args
25      */
26     public static void main(String[] args) {
27         // TODO Auto-generated method stub
28         Solution2 sol = new Solution2();
29         List<Integer> res = sol.maxCycleLen(1, 100000);
30         System.out.println(res);
31     }
32
33 }

follow up: 建立一个Lookup table, 算过的数就不算了

 1 package Sorting;
 2 import java.util.*;
 3
 4 public class Solution3 {
 5     HashMap<Integer, Integer> map;
 6     public List<Integer> maxCycleLen(int min, int max) {
 7         List<Integer> maxCycle = new ArrayList<Integer>();
 8         map = new HashMap<Integer, Integer>();
 9         for (int i=min; i<=max; i++) {
10             helper(maxCycle, i);
11         }
12         return maxCycle;
13     }
14
15     public void helper(List<Integer> maxCycle, int num) {
16         int len = 1;
17         int numcopy = num;
18         while (num != 1) {
19             if (map.containsKey(num)) {
20                 len += map.get(num)-1;
21                 break;
22             }
23             if (num%2 == 1) num = num*3+1;
24             else num = num/2;
25             len++;
26         }
27         maxCycle.add(len);
28         map.put(numcopy, len);
29     }
30
31     /**
32      * @param args
33      */
34     public static void main(String[] args) {
35         // TODO Auto-generated method stub
36         Solution3 sol = new Solution3();
37         List<Integer> res = sol.maxCycleLen(1, 100000);
38         System.out.println(res);
39     }
40
41 }

或者

 1     public void helper(List<Integer> maxCycle, int num) {
 2         int len = 0;
 3         int numcopy = num;
 4         while (!map.containsKey(num)) {
 5             if (num == 1) {
 6                 map.put(1, 1);
 7                 maxCycle.add(1);
 8                 return;
 9             }
10             if (num%2 == 1) num = num*3+1;
11             else num = num/2;
12             len++;
13         }
14         len = len + map.get(num);
15         maxCycle.add(len);
16         map.put(numcopy, len);
17     }