二刷 06/2022
Version #1 Choose the next number and add the result at each layer
Time O(N* 2^N)
Space O (N)
Runtime: 2 ms, faster than 76.48% of Java online submissions for Subsets II.
Memory Usage: 43.8 MB, less than 63.83% of Java online submissions for Subsets II.
class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
if (nums == null || nums.length == 0) {
return result;
}
Arrays.sort(nums);
dfs(nums, 0, new ArrayList<>(), result);
return result;
}
private void dfs(int[] nums, int startIndex, List<Integer> path, List<List<Integer>> result) {
result.add(new ArrayList<>(path));
for (int i = startIndex; i < nums.length; i++) {
// For the same layer, only add duplicate element once
if (i != startIndex && nums[i] == nums[i - 1]) {
continue;
}
path.add(nums[i]);
dfs(nums, i + 1, path, result);
path.remove(path.size() - 1);
}
}
}
一刷
Version #2 [TODO]decide 1 number chosen or not at each layer
Version #1 Choosing the next number, add to result at each layer
71.00 %
public class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
if (nums == null || nums.length == 0) return result;
Arrays.sort(nums);
subsetsWithDupDfsHelper(nums, 0, new ArrayList<Integer>(), result);
return result;
}
private void subsetsWithDupDfsHelper(int[] nums, int startIndex, List<Integer> path, List<List<Integer>> result) {
result.add(new ArrayList<Integer>(path));
for (int i = startIndex; i < nums.length; i++) {
if (i != startIndex && nums[i] == nums[i - 1]) continue;
path.add(nums[i]);
subsetsWithDupDfsHelper(nums, i + 1, path, result);
path.remove(path.size() - 1);
}
}
}
No comments:
Post a Comment