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Update 90. Subsets II.md

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@ -17,7 +17,8 @@
/ \ / \ / \ / \
[1 2 2] [1 2] X [1] [2 2] [2] X []
```
需要注意去掉重复值，即对树进行剪枝: 当前一层的元素（`nums[level-1]`）未被访问且`nums[level]==nums[level-1]`时，当前元素`nums[level]`也不应该访问，即只沿右子树下降。
需要注意避免重复结果: 如果下一层对应元素满足`nums[level+1] == nums[level]`，若不打算访问当前层对应元素（`nums[level]`），那么也不应该访问下一层对应元素，所以在向右子树下降之前应该跳过重复值，详见代码。
## 思路二
此题也可以使用迭代的方式来完成此题。
参考[78. Subsets题解](https://github.com/ShusenTang/LeetCode/blob/master/solutions/78.%20Subsets.md)，当处理到第一个2时，此时的子集合为[], [1], [2], [1, 2]，
@ -32,33 +33,28 @@
``` C++
class Solution {
private:
void DFS(vector<vector<int>>&res, vector<int>&subset, vector<bool>&visited, const vector<int>& nums, int level){
if(level == nums.size()) { // 叶节点
res.push_back(subset);
void DFS(const vector<int>&nums, vector<vector<int>>&res, vector<int>&out, int level){
if(level == nums.size()){
res.push_back(out);
return;
}
if(level > 0 && nums[level] == nums[level - 1] && !visited[level - 1]){
DFS(res, subset, visited, nums, level + 1); // 沿右子树下降
return;
}
// 左子树
out.push_back(nums[level]);
DFS(nums, res, out, level+1);
out.pop_back();
// 沿左子树下降
visited[level] = true;
subset.push_back(nums[level]);
DFS(res, subset, visited, nums, level + 1);
visited[level] = false;
subset.pop_back();
// 沿右子树下降
DFS(res, subset, visited, nums, level + 1);
// 右子树
int i = level + 1;
while(i < nums.size() && nums[i] == nums[level]) i++; // 避免重复, 唯一和78. Subsets不同的地方
DFS(nums, res, out, i);
}
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
vector<vector<int>>res;
vector<int>subset;
vector<bool>visited(nums.size(), false);
vector<int>out;
sort(nums.begin(), nums.end());
DFS(res, subset, visited, nums, 0);
DFS(nums, res, out, 0);
return res;
}
};