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101 lines (83 loc) · 2.71 KB
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package practice.recursion;
import common.ArrayUtil;
import java.util.ArrayList;
import java.util.List;
/**
* Given an integer array nums, return all possible subsets (the power set).
The solution set must not contain duplicate subsets.
Example 1:
Input: nums = [1,2,3]
Output: [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]
Example 2:
Input: nums = [0]
Output: [[],[0]]
Constraints:
1 <= nums.length <= 10
-10 <= nums[i] <= 10
All the numbers of nums are unique.
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/subsets
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*/
public class SubSets {
private List<List<Integer>> res;
/**
* 利用二进制位 将 2^n个结果 依次展开，循环loop获得结果
* @param nums
* @return
*/
public List<List<Integer>> loop(int[] nums) {
int total = 1 << nums.length;
res = new ArrayList<>();
for (int i = 0; i < total; i++) {
List<Integer> row = new ArrayList<>();
for (int j = 0; j < nums.length; j++)
if (((i >> (nums.length - j - 1)) & 1) != 0) row.add(nums[j]);
res.add(row);
}
return res;
}
public List<List<Integer>> loop2(int[] nums) {
res = new ArrayList<>();
res.add(new ArrayList<>());
for (int n : nums) {
List<List<Integer>> newSubsets = new ArrayList<>();
for (List<Integer> subset : res) {
subset.add(n);
newSubsets.add(new ArrayList<>(subset));
subset.remove(subset.size() - 1);
}
res.addAll(newSubsets);
}
return res;
}
/**
* 经典分治法，分解为子问题: 在每一位上分别考虑保留/舍弃
* @param nums
* @return
*/
public List<List<Integer>> s1(int[] nums) {
res = new ArrayList<>();
recursion(nums, 0, new ArrayList<>());
return res;
}
public void recursion(int[] nums, int pos, List<Integer> row) {
if (pos >= nums.length) {
res.add(new ArrayList<>(row));
return;
}
recursion(nums, pos + 1, row);
row.add(nums[pos]);
recursion(nums, pos + 1, row);
row.remove(row.size() - 1);
}
public static void main(String[] args) {
int[] arr = new int[]{1,2,3};
SubSets sol = new SubSets();
List<List<Integer>> res = sol.loop(arr);
List<List<Integer>> res1 = sol.s1(arr);
System.out.printf("> Input: %s\n", ArrayUtil.toString(arr));
System.out.printf("> Output: \nloop=%s\nrecursion=%s\n", res.toString(), res1.toString());
}
}