代码随想录算法训练营第29天|491.递增子序列 * * 46.全排列 * 47.全排列 II

491.递增子序列

思路：

代码

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> path=new LinkedList<>();
    public List<List<Integer>> findSubsequences(int[] nums) {
        backtracking(nums,0);
        return res;
    }
    public void backtracking(int[] nums,int startIndex){
        // if(startIndex>=nums.length-1)return;
        if(path.size()>1){
            res.add(new ArrayList<>(path));
        }
        int[] used = new int[201];
        for(int i=startIndex;i<nums.length;i++){
            if(!path.isEmpty()&&path.getLast()> nums[i]|| (used[nums[i]+100]==1)){
                continue;
            }
            used[nums[i]+100]=1;
            path.add(nums[i]);
            backtracking(nums,i+1);
            path.removeLast();
        }
    }
}

思路：优化

通过数组去重，取代hashset

代码：

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> path=new ArrayList<>();
    public List<List<Integer>> findSubsequences(int[] nums) {
        backtracking(nums,0);
        return res;
    }
    public void backtracking(int[] nums,int startIndex){
        // if(startIndex>=nums.length-1)return;
        if(path.size()>1){
            res.add(new ArrayList<>(path));
        }
        int[] used = new int[201];
        for(int i=startIndex;i<nums.length;i++){
            if(!path.isEmpty()&&path.get(path.size() -1 ) > nums[i]|| (used[nums[i]+100]!=0)){
                continue;
            }
            used[nums[i]+100]=1;
            path.add(nums[i]);
            backtracking(nums,i+1);
            path.remove(path.size() -1);
        }
    }
}

46.全排列

思路

代码一：使用used数组

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    public List<List<Integer>> permute(int[] nums) {
        boolean[] used=new boolean[nums.length];
        backtracking(nums,used);
        return res;
    }
    public void backtracking(int[] nums,boolean[] used){
        if(path.size()==nums.length){
            res.add(new ArrayList<>(path));
            return;
        }
        
        for(int i=0;i<nums.length;i++){
            // System.out.println("数值"+nums[i]+"布尔值"+used[nums[i]+10]);
            if(used[i]){
                continue;
            }
            path.add(nums[i]);
            used[i]=true;
            backtracking(nums,used);
            used[i]=false;
            path.remove(path.size()-1);
        }
    }
}

代码二：使用path判断元素

path是linkedlist

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> path = new LinkedList<>();
    public List<List<Integer>> permute(int[] nums) {
        // boolean[] used=new boolean[nums.length];
        backtracking(nums);
        return res;
    }
    public void backtracking(int[] nums){
        if(path.size()==nums.length){
            res.add(new ArrayList<>(path));
            return;
        }
        
        for(int i=0;i<nums.length;i++){
            // System.out.println("数值"+nums[i]+"布尔值"+used[nums[i]+10]);
            if(path.contains(nums[i])){
                continue;
            }
            path.add(nums[i]);
            backtracking(nums);
            path.removeLast();
        }
    }
}

47.全排列 II

思路一：层节点和路径都是用used数组做记录

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    public List<List<Integer>> permuteUnique(int[] nums) {
        boolean[] used=new boolean[nums.length];
        backtracking(nums,used);
        return res;
    }
    public void backtracking(int[] nums,boolean[] used){
        if(path.size()==nums.length){
            res.add(new ArrayList<>(path));
            return;
        }
        int[] cengused=new int[21];
        for(int i=0;i<nums.length;i++){
            // System.out.println("数值"+nums[i]+"布尔值"+used[nums[i]+10]);
            if(used[i]||cengused[nums[i]+10]==1){
                continue;
            }
            path.add(nums[i]);
            cengused[nums[i]+10]=1;
            used[i]=true;
            backtracking(nums,used);
            used[i]=false;
            path.remove(path.size()-1);
        }
    }
}

思路二：层通过排序后是否重复过滤

class Solution {
    //存放结果
    List<List<Integer>> result = new ArrayList<>();
    //暂存结果
    List<Integer> path = new ArrayList<>();

    public List<List<Integer>> permuteUnique(int[] nums) {
        boolean[] used = new boolean[nums.length];
        Arrays.fill(used, false);
        Arrays.sort(nums);
        backTrack(nums, used);
        return result;
    }

    private void backTrack(int[] nums, boolean[] used) {
        if (path.size() == nums.length) {
            result.add(new ArrayList<>(path));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            // used[i - 1] == true，说明同⼀树⽀nums[i - 1]使⽤过
            // used[i - 1] == false，说明同⼀树层nums[i - 1]使⽤过
            // 如果同⼀树层nums[i - 1]使⽤过则直接跳过
            if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) {
                continue;
            }
            //如果同⼀树⽀nums[i]没使⽤过开始处理
            if (used[i] == false) {
                used[i] = true;//标记同⼀树⽀nums[i]使⽤过，防止同一树枝重复使用
                path.add(nums[i]);
                backTrack(nums, used);
                path.remove(path.size() - 1);//回溯，说明同⼀树层nums[i]使⽤过，防止下一树层重复
                used[i] = false;//回溯
            }
        }
    }
}