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
package Sorting;
import java.util.*; public class Solution2 {
public List<Integer> maxCycleLen(int min, int max) {
List<Integer> maxCycle = new ArrayList<Integer>();
for (int i=min; i<=max; i++) {
helper(maxCycle, i);
}
return maxCycle;
} public void helper(List<Integer> maxCycle, int num) {
int len = 1;
while (num != 1) {
if (num%2 == 1) num = num*3+1;
else num = num/2;
len++;
}
maxCycle.add(len);
} /**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Solution2 sol = new Solution2();
List<Integer> res = sol.maxCycleLen(1, 100000);
System.out.println(res);
} }
follow up: 建立一个Lookup table, 算过的数就不算了
package Sorting;
import java.util.*; public class Solution3 {
HashMap<Integer, Integer> map;
public List<Integer> maxCycleLen(int min, int max) {
List<Integer> maxCycle = new ArrayList<Integer>();
map = new HashMap<Integer, Integer>();
for (int i=min; i<=max; i++) {
helper(maxCycle, i);
}
return maxCycle;
} public void helper(List<Integer> maxCycle, int num) {
int len = 0;
int numcopy = num;
while (!map.containsKey(num)) {
if (num == 1) {
map.put(1, 1);
maxCycle.add(1);
return;
}
if (num%2 == 1) num = num*3+1;
else num = num/2;
len++;
}
len = len + map.get(num);
maxCycle.add(len);
map.put(numcopy, len);
} /**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Solution3 sol = new Solution3();
List<Integer> res = sol.maxCycleLen(1, 100000);
System.out.println(res);
} }
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