LeetCode OJ - Subsets II

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Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.

For example,
If S = [1,2,2], a solution is:

[  [2],  [1],  [1,2,2],  [2,2],  [1,2],  []]

分析：如果仍用subsets的方法来做，里面会有重复的item，例如010和001的到的item都为[2]

方法一：subsets改进

DFS(s, layer, item) {    if(layer == s.size()) {        for(int i = 0; i < ret.size(); i++) {            if(ret.at(i) == item) return;    //此处也可以用ret[i];        }        ret += item;    }    DFS(s, layer+1, item);    item.push_back(s[layer]);    DFS(s, layer+1, item);}

方法二：以{1、2、3}为例,根据空子集 + {1...、 2...、 3... }来构建树，每一次函数调用访问树的所有子节点。然后去重

ret += path;  //空子集DFS(s, start, path) {    for(i = start; i < s.size(); i++) {        if(i != start && s[i] == s[i-1]) continue;        path += s[i];        ret += path;        DFS(s, i+1; path);        path -= s[i];    }}

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