回溯算法part05 491. 非递减子序列 46. 全排列 47. 全排列 II
class Solution {
private:vector<vector<int>> result;vector<int> path;
public:void backTracing(vector<int>& nums,int startIndex){if(path.size()>1) result.push_back(path);unordered_set<int> uset;for(int i = startIndex;i<nums.size();i++){if( (!path.empty() && nums[i]<path.back()) || uset.find(nums[i]) != uset.end()) continue; //非递增数组或者单层元素重复,跳过该层元素uset.insert(nums[i]); // 记录这个元素在本层用过了,本层后面不能再用了path.push_back(nums[i]);backTracing(nums,i+1);path.pop_back();}}vector<vector<int>> findSubsequences(vector<int>& nums) {backTracing(nums,0);return result;}
};
class Solution {
private:vector<vector<int>> result;vector<int> path;
public:void backtracking(vector<int>& nums,vector<int>& used){ //值得注意的是,这里面的used是深度遍历的,与491本层使用有差别,所以去重位置有区别if(path.size()==nums.size()){result.push_back(path);return;} //叶子节点回收元素for(int i = 0; i<nums.size();i++){if(used[i]==1) continue;used[i] = 1;path.push_back(nums[i]);backtracking(nums,used);path.pop_back();used[i] = 0;}}vector<vector<int>> permute(vector<int>& nums) {vector<int> used(nums.size(),0);backtracking(nums,used);return result;}
};
class Solution {
public:vector<vector<int>> result;vector<int> path;void backtracking(vector<int>& nums,vector<int>& used){// 此时说明找到了一组if (path.size() == nums.size()) {result.push_back(path);return;}for(int i = 0;i<nums.size();i++){if(i>0&&nums[i]==nums[i-1]&&used[i - 1] ==0) continue; //used[i - 1] ==0至关重要,树层上去重if(!used[i]){used[i] = 1;path.push_back(nums[i]);backtracking(nums,used);used[i] = 0;path.pop_back();}}}vector<vector<int>> permuteUnique(vector<int>& nums) {sort(nums.begin(), nums.end()); //排序vector<int> used(nums.size(),0);backtracking(nums,used);return result;}
};