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Leetcode - Subsets I，II
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Leetcode - 078. Subsets
這道題重定義了什么叫可行解：
一般而言，可行解需要滿足強約束性條件集，而本題的可行解就是單一弱約束性條件(distinct integers,只需要當前集合內的元素不重復即可算作一個可行解) ，算是比較簡單的入門級dfs + backtracing 題目。
class Solution {
public:void dfs(vector<vector<int>> & vct,vector<int> &cur,vector<int>& nums,int index){vct.push_back(cur);int n = nums.size();if(index >= n)return;for(int i = index;i < n;++i){cur.push_back(nums[i]);dfs(vct,cur,nums,i + 1);cur.pop_back();}}vector<vector<int>> subsets(vector<int>& nums) {vector<vector<int>> vct;vector<int> cur;dfs(vct,cur,nums,0);return vct;}
};
Leetcode - 090. Subsets II
diff vs the pre version:
有重復的元素怎么算？相關的思路雷同于Leetcode - 040. Combination Sum II
class Solution {
public:void dfs(vector<vector<int>> &vct, vector<int> &cur, vector<int>& nums,vector<int> & used,int index){vct.push_back(cur);int n = nums.size();if (index >= n)return;for (int i = index; i < n; ++i){if (used[i] == 0){int j = i - 1;bool repeated = false;while (j >= 0 && nums[j] == nums[i]){if (used[j] == 0){repeated = true;break;}--j;}if (repeated)continue;cur.push_back(nums[i]);used[i] = 1;dfs(vct, cur, nums, used, i + 1);used[i] = 0;cur.pop_back();}}}vector<vector<int>> subsetsWithDup(vector<int>& nums) {vector<vector<int>> vct;vector<int> cur;int n = nums.size();if (n <= 0)return vct;vector<int> used(n, 0);sort(nums.begin(), nums.end());dfs(vct, cur, nums,used,0);return vct;}
};
總結
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