Open azl397985856 opened 2 weeks ago
时间复杂度:O(n*2^n) 空间On
public List<List<Integer>> subsets(int[] nums) {
getZiji(nums,0,new ArrayList<>());
return res;
}
List<List<Integer>> res = new ArrayList<>();
private void getZiji(int[] nums,int start,List<Integer> track){
res.add(new ArrayList<>(track));
for(int i = start;i<nums.length;i++){
track.add(nums[i]);
getZiji(nums,i+1,track);
track.remove(track.size()-1);
}
}
我通过递归实现的
/*
* @lc app=leetcode.cn id=78 lang=javascript
*
* [78] 子集
*/
// @lc code=start
/**
* @param {number[]} nums
* @return {number[][]}
*/
var subsets = function (nums) {
const select = [];
const ans = [];
const dfs = (index) => {
if (index === nums.length) {
ans.push([...select]);
return;
}
select.push(nums[index]);
dfs(index + 1);
select.pop();
dfs(index + 1);
};
dfs(0);
return ans;
};
// @lc code=end
class Solution:
def subsets(self, nums: List[int]) -> List[List[int]]:
rs = [[]]
for n in nums:
tmp = []
for r in rs:
tmp.append(r+[n])
rs+=tmp
return rs
class Solution:
def subsets(self, nums: List[int]) -> List[List[int]]:
ans = []
for i in range(2**len(nums)):
cur = []
for j in range(len(nums)):
if ((1 << j) & i) != 0:
cur.append(nums[j])
ans.append(cur)
return ans
78. 子集
入选理由
暂无
题目地址
https://leetcode-cn.com/problems/subsets/
前置知识
题目描述
解集 不能 包含重复的子集。你可以按 任意顺序 返回解集。
示例 1:
输入:nums = [1,2,3] 输出:[[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]] 示例 2:
输入:nums = [0] 输出:[[],[0]]
提示:
1 <= nums.length <= 10 -10 <= nums[i] <= 10 nums 中的所有元素 互不相同