【发布时间】:2014-12-05 04:56:42
【问题描述】:
以下是用于生成多个集合的算法。正好解决了下面的问题Print all combination of element in the array such that first element of array is d and
next element in the array can be +1 or -1 the previous element in the array. Code was required.
Input: num=4, level=3. Output - 4, 3, 2 || 4, 3, 4 || 4, 5, 4 || 4, 5, 6。
这个问题没有使用 int[] arr = new int[level] 作为数据存储,它最终被复制到一个 int 数组列表中。
以下代码的空间复杂度是多少?是 O(level) 即用于解决问题的存储还是 O(2^level - 1) 即返回类型的大小?
public static List<int[]> getCombinations(int num, int level) {
final List<int[]> combinations = new ArrayList<int[]>();
int[] arr = new int[level];
computeCombinations(num, level - 1, 0, combinations, arr); // note : its level - 1, since currentlevel is set to 0.
return combinations;
}
private static void computeCombinations(int num, int level, int currLevel, List<int[]> list, int[] arr) {
arr[currLevel] = num;
if (currLevel == level) {
// list.add(arr); <- wrong to do it so
int[] temp = new int[arr.length];
System.arraycopy(arr, 0, temp, 0, arr.length);
list.add(temp);
return;
}
computeCombinations(num - 1, level, currLevel + 1, list, arr);
computeCombinations(num + 1, level, currLevel + 1, list, arr);
}
【问题讨论】:
-
我会说
O(level * 2^(level-1)),因为代码显然会在内存中生成整个列表。
标签: java algorithm complexity-theory space-complexity