Leetcode C++《热题 Hot 100-42》78.子集
- 题目
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。
示例:
输入: nums = [1,2,3]
输出:
[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/subsets
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- 思路
- 方案1:相关标签,看到了位运算,还在想怎么结合,因为每一个值可以选或者不能选
- 时间复杂度(1 << nums.size())
- 方案2:递归回溯做法, 递归的基本思路,从i个元素到i+1个元素的过程
- 代码
class Solution {
public:
//方案1
/*vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> res;
int subsetNum = 1 << nums.size();
for(unsigned int i = 0; i < subsetNum; i++) { //循环8的
vector<int> oneres;
for (int j = 0; j < nums.size(); j++) {
if (((i>>j)&1) == 1) {
oneres.push_back(nums[j]);
}
}
res.push_back(oneres);
}
return res;
}*/
//方案2:
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> res;
if (nums.size() == 0)
return res;
backTracking(nums, 0, res);
res.push_back(vector<int>());
return res;
}
void backTracking(vector<int> nums, int add_index, vector<vector<int>>& subsets) {
if (add_index == nums.size())
return;
vector<vector<int>> newSubsets;
vector<int> oneSubset;
for (int i = 0; i < subsets.size(); i++) {
oneSubset.clear();
oneSubset = subsets[i];
oneSubset.push_back(nums[add_index]);
newSubsets.push_back(oneSubset);
}
oneSubset.clear();
oneSubset.push_back(nums[add_index]);
newSubsets.push_back(oneSubset);
for (int i = 0; i < newSubsets.size(); i++) {
subsets.push_back(newSubsets[i]);
}
backTracking(nums, add_index+1, subsets);
}
};