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}}}
}