0078. Subsets (M)

Subsets (M)

题目

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: nums = [1,2,3]
Output:
[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

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题意

求一个数组的幂集(即所有子集，包括全集和空集，构成的集族)。

思路

回溯法。

也可以利用位运算来求子集，具体方法参考 LeetCode 78. Subsets （位运算入门：利用位运算求子集）。

也可以直接循环迭代处理。

---

代码实现

Java

回溯法

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> ans = new ArrayList<>();
        subsets(nums, 0, new ArrayList<>(), ans);   
        return ans;
    }

    private void subsets(int[] nums, int index, List<Integer> list, List<List<Integer>> ans) {
        if (index == nums.length) {
            ans.add(new ArrayList<>(list));
            return;
        }

        subsets(nums, index + 1, list, ans);
        list.add(nums[index]);
        subsets(nums, index + 1, list, ans);
        list.remove(list.size() - 1);
    }
}

位运算

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> ans = new ArrayList<>();
        List<Integer> list = new ArrayList<>();
        int all = 1 << (nums.length);		// 总情况数
        for (int i = 0; i < all; i++) {
            // 判断数组中每一个数是否应该添加
            for (int j = 0; j < nums.length; j++) {
                if ((i & (1 << j)) > 0) {
                    list.add(nums[j]);
                }
            }
            ans.add(new ArrayList<>(list));
            list.clear();
        }
        return ans;
    }
}

JavaScript

回溯法

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var subsets = function (nums) {
  let lists = []
  dfs(nums, 0, [], lists)
  return lists
}

let dfs = function (nums, index, list, lists) {
  if (index === nums.length) {
    lists.push([...list])
    return
  }

  dfs(nums, index + 1, list, lists)
  list.push(nums[index])
  dfs(nums, index + 1, list, lists)
  list.pop()
}

位运算

	/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var subsets = function (nums) {
  let lists = []
  let count = 1 << nums.length

  for (let i = 0; i < count; i++) {
    let list = []
    for (let j = 0; j < nums.length; j++) {
      if (i & (1 << j)) {
        list.push(nums[j])
      }
    }
    lists.push([...list])
  }

  return lists
}

迭代

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var subsets = function (nums) {
  let lists = [[]]
  for (let i = 0; i < nums.length; i++) {
    let size = lists.length
    for (let j = 0; j < size; j++) {
      lists.push(lists[j].concat(nums[i]))
    }
  }
  return lists
}