LeetCode OJ-78.Subsets
題目描述
Given a set of distinct integers, nums, return all possible subsets.
Note: The solution set must not contain duplicate subsets.
For example,
If nums =[1,2,3]
, a solution is:
[ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
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題目理解
題目大意是要求解出一個給定序列的所有子集,即長度爲0,1,…n的所有子集,n爲給定序列的長度。
這題思想和LeetCode 77類似,使用回溯法可以解決,同樣,也要注意去除重複的情況,[1,2]與[2,1]是一樣的,這種要剔除。這裏由於給定序列裏元素是不定的,所以先要排序,之後使用LeetCode 77裏的處理方法,每次遍歷選擇元素時,從i(i = m)到n,之後將nums[i]壓入vector,隨後以i + 1進入遞歸,遍歷選擇元素從i + 1到n。這樣就可以保證前面選擇到的元素,後面將不會被選中。
subsets中要利用一個循環,以子集長度爲循環條件,從0一直選擇到n。具體代碼如下:
Code
void backtrace(int n, int m, int k, int l, const vector<int> &nums,
vector<int> &rec, vector<vector<int>> &res)
{
if (l == k) {
res.push_back(rec);
return ;
}
else if (m >= n) {
return ;
}
int i;
for (i = m; i < n; ++i) {
rec.push_back(nums[i]);
backtrace(n, i + 1, k, l + 1, nums, rec, res);
rec.pop_back();
}
}
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> res;
vector<int> rec;
sort(nums.begin(), nums.end());
int i;
int sz = (int) nums.size();
for (i = 0; i <= sz; ++i) {
backtrace(sz, 0, i, 0, nums, rec, res);
}
return res;
}
};