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105 lines
2.1 KiB
Markdown
Executable File
105 lines
2.1 KiB
Markdown
Executable File
# [78. Subsets](https://leetcode.com/problems/subsets/)
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## 题目
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Given a set of **distinct** integers, *nums*, return all possible subsets (the power set).
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**Note**: The solution set must not contain duplicate subsets.
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**Example**:
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Input: nums = [1,2,3]
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Output:
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[
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[3],
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[1],
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[2],
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[1,2,3],
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[1,3],
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[2,3],
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[1,2],
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[]
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]
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## 题目大意
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给定一组不含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。说明：解集不能包含重复的子集。
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## 解题思路
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- 找出一个集合中的所有子集，空集也算是子集。且数组中的数字不会出现重复。用 DFS 暴力枚举即可。
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- 这一题和第 90 题，第 491 题类似，可以一起解答和复习。
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## 代码
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```go
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package leetcode
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import "sort"
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// 解法一
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func subsets(nums []int) [][]int {
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c, res := []int{}, [][]int{}
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for k := 0; k <= len(nums); k++ {
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generateSubsets(nums, k, 0, c, &res)
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}
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return res
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}
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func generateSubsets(nums []int, k, start int, c []int, res *[][]int) {
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if len(c) == k {
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b := make([]int, len(c))
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copy(b, c)
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*res = append(*res, b)
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return
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}
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// i will at most be n - (k - c.size()) + 1
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for i := start; i < len(nums)-(k-len(c))+1; i++ {
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c = append(c, nums[i])
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generateSubsets(nums, k, i+1, c, res)
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c = c[:len(c)-1]
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}
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return
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}
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// 解法二
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func subsets1(nums []int) [][]int {
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res := make([][]int, 1)
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sort.Ints(nums)
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for i := range nums {
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for _, org := range res {
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clone := make([]int, len(org), len(org)+1)
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copy(clone, org)
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clone = append(clone, nums[i])
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res = append(res, clone)
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}
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}
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return res
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}
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// 解法三：位运算的方法
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func subsets2(nums []int) [][]int {
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if len(nums) == 0 {
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return nil
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}
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res := [][]int{}
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sum := 1 << uint(len(nums))
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for i := 0; i < sum; i++ {
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stack := []int{}
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tmp := i // i 从 000...000 到 111...111
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for j := len(nums) - 1; j >= 0; j-- { // 遍历 i 的每一位
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if tmp & 1 == 1 {
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stack = append([]int{nums[j]}, stack...)
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}
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tmp >>= 1
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}
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res = append(res, stack)
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}
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return res
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}
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``` |