Difficulty:: Medium
Given a set of distinct integers, nums, return all possible subsets (the power set).
Note: The solution set must not contain duplicate subsets.
Example:
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| Input: nums = [1,2,3]
Output:
[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]
|
Solution
Language: Java
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| class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
if (nums == null) {
return result;
}
dfsHelper(nums, result, new ArrayList<Integer>(), 0);
return result;
}
private void dfsHelper(int[] nums, List<List<Integer>> result, List<Integer> subset, int index) {
result.add(new ArrayList<Integer>(subset));
for (int i = index; i < nums.length; i++) {
subset.add(nums[i]);
dfsHelper(nums, result, subset, i + 1);
subset.remove(subset.size() - 1);
}
}
}
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使用位运算
一共有$2^n$种可能性，如果用一个32进制数来确定一个数是否存在，就直接遍历0到n位全1的数即可。这种情况建立在nums不会超过31个的基础上，因为数字个数多了的话子集数太大，所以一般不会超过这个数字
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| class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
if (nums == null) {
return result;
}
int len = nums.length;
int max = 1;
for (int i = 0; i < len - 1; i++) {
max = (max << 1) + 1;
}
for (int i = 0; i <= max; i++) {
List<Integer> subset = new ArrayList<>();
int tmp = i;
for (int j = 0; j < len; j++) {
if ((tmp & 1) != 0) {
subset.add(nums[j]);
}
tmp >>= 1;
}
result.add(subset);
}
return result;
}
}
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