下面是用於生成多個集合的算法。正是它解決了以下問題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
。是否爲空間複雜度返回值帳戶
此問題未使用int[] arr = new int[level]
作爲最終被複制到int數組列表中的數據存儲。
以下代碼的空間複雜度是多少?它是O(級別),即用於解決問題的存儲,還是O(2^level-1),即返回類型的大小?
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);
}
我會說'O(level * 2 ^(level-1))',因爲代碼顯然會在內存中生成整個列表。 – user3386109 2014-12-05 05:14:49