标签：排列 22 nums 46 int used new path backtraking
491.递增子序列
491. 非递减子序列
class Solution {
public List<Integer> path=new LinkedList<>();
public List<List<Integer>> res=new ArrayList<>();
public List<List<Integer>> findSubsequences(int[] nums) {
backtraking(nums,0);
return res;
}
void backtraking(int[] nums,int startIndex)
{
if(path.size()>=2)
{
res.add(new ArrayList<>(path));
}
HashSet<Integer> hs=new HashSet<>();
for(int i=startIndex;i<nums.length;i++)
{
//去重，树层不能重复
if(!path.isEmpty()&&path.get(path.size()-1)>nums[i]|| hs.contains(nums[i])) continue;
hs.add(nums[i]);
path.add(nums[i]);
backtraking(nums,i+1);
path.remove(path.size()-1);
}
}
}
46.全排列
46. 全排列
class Solution {
public List<Integer> path=new LinkedList<>();
public List<List<Integer>> res=new ArrayList<>();
public List<List<Integer>> permute(int[] nums) {
int[] used=new int[nums.length];
backtraking(nums,used);
return res;
}
void backtraking(int[] nums,int[] used)
{
if (path.size()==nums.length)
{
res.add(new ArrayList<>(path));
return;
}
for(int i=0;i< nums.length;i++)
{
if(used[i]==0)
{
path.add(nums[i]);
used[i]=1;
backtraking(nums,used);
path.remove(path.size()-1);
used[i]=0;
}
}
}
}
47.全排列 II
47. 全排列 II
class Solution {
public List<Integer> path=new LinkedList<>();
public List<List<Integer>> res=new ArrayList<>();
public List<List<Integer>> permuteUnique(int[] nums) {
Arrays.sort(nums);
int[] used=new int[nums.length];
backtraking(nums,used);
return res;
}
void backtraking(int[] nums,int[] used)
{
if (path.size()==nums.length)
{
res.add(new ArrayList<>(path));
return;
}
for(int i=0;i< nums.length;i++)
{
if(i>0&&nums[i]==nums[i-1]&&used[i-1]==0)continue;
if(used[i]==0)
{
path.add(nums[i]);
used[i]=1;
backtraking(nums,used);
path.remove(path.size()-1);
used[i]=0;
}
}
}
}
332.重新安排行程 (一刷跳过)
51.N皇后（一刷跳过）
37.解数独（一刷跳过）
标签：排列,
22,
nums,
46,
int,
used,
new,
path,
backtraking
From： https://www.cnblogs.com/FreeDrama/p/18425798