Groupon面经Prepare: Max Cycle Length - neverlandly - 博客园

Groupon面经Prepare: Max Cycle Length - neverlandly - 博客园

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

 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     }

follow up: 建立一个Lookup table, 算过的数就不算了

 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 = 0;
17         int numcopy = num;
18         while (!map.containsKey(num)) {
19             if (num == 1) {
20                 map.put(1, 1);
21                 maxCycle.add(1);
22                 return;
23             }
24             if (num%2 == 1) num = num*3+1;
25             else num = num/2;
26             len++;
27         }
28         len = len + map.get(num);
29         maxCycle.add(len);
30         map.put(numcopy, len);
31     }

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