子集和排列

分类: IT文章 • 2022-04-02 19:06:39

子集和排列

1 Subsets

    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums.length == 0) {
            return result;
        }
        ArrayList<Integer> list = new ArrayList<>();
        subsetsHelper(nums, 0, list, result);
        return result;
    }
    void subsetsHelper(int[] nums, int pos, List<Integer> list, List<List<Integer>> result){
        result.add(new ArrayList<>(list));
        for (int i = pos; i < nums.length; i++) {
            list.add(nums[i]);
            subsetsHelper(nums, i + 1, list, result);
            list.remove(list.size() - 1);
        }
    }

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2 Subsets II

    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums.length == 0) {
            return result;
        }
        Arrays.sort(nums);
        ArrayList<Integer> list = new ArrayList<>();
        subsetsHelper(nums, 0, list, result);
        return result;
    }
    void subsetsHelper(int[] nums, int pos, List<Integer> list, List<List<Integer>> result){
        result.add(new ArrayList<>(list));
        for (int i = pos; i < nums.length; i++) {
            if (i > pos && nums[i] == nums[i - 1]){
                continue;
            }
            list.add(nums[i]);
            subsetsHelper(nums, i + 1, list, result);
            list.remove(list.size() - 1);
        }
    }

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3 Permutations

    public List<List<Integer>> permute(int[] nums) {
        // write your code here
        List<List<Integer>> res = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            res.add(new ArrayList<Integer>());
            return res;
        }
        List<Integer> list = new ArrayList<>();
        helper(nums, list, res);
        return res;
    }
    
    void helper(int[] nums, List<Integer> list, List<List<Integer>> res) {
        if (list.size() == nums.length) {
            res.add(new ArrayList<>(list));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            if (list.contains(nums[i])) {
                continue;
            }
            list.add(nums[i]);
            helper(nums, list, res);
            list.remove(list.size() - 1);
        }
    }

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4 Permutations II

    public List<List<Integer>> permuteUnique(int[] nums) {
        // Write your code here
        List<List<Integer>> result = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            result.add(new ArrayList<Integer>());
            return result;
        }
        Arrays.sort(nums);
        List<Integer> list = new ArrayList<>();
        boolean[] used = new boolean[nums.length];
        helper(nums, list, used, result);
        return result;
    }
    void helper(int[] nums, List<Integer> list, boolean[] used, List<List<Integer>> result) {
        if (list.size() == nums.length) {
            result.add(new ArrayList<>(list));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            if (used[i] || (i > 0 && nums[i] == nums[i - 1] && !used[i - 1])) {
                continue;
            }
            used[i] = true;
            list.add(nums[i]);
            helper(nums, list, used, result);
            used[i] = false;
            list.remove(list.size() - 1);
        }
    }

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5 Combination Sum 可以有重复

    public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) 
    {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if (candidates == null || candidates.length == 0)
        {
            return res;
        }
        Arrays.sort(candidates);
        helper(candidates, 0, target, new ArrayList<Integer>(), res);
        return res;
    }
    private void helper(int[] candidates, int start, int target, ArrayList<Integer> item,
            ArrayList<ArrayList<Integer>> res) {
        // TODO Auto-generated method stub
        if (target < 0)
        {
            return;
        }
        if (target == 0)
        {
            res.add(new ArrayList<>(item));
            return;
        }
        for (int i = start; i < candidates.length; i++)
        {
            if (i > 0 && candidates[i] == candidates[i - 1])
            {
                continue;
            }
            item.add(candidates[i]);
            helper(candidates, i, target - candidates[i], item, res);
            item.remove(item.size() - 1);
        }
    }

View Cod

6 Combination Sum 不可以有重复

public List<List<Integer>> combinationSum2(int[] nums, int target) {
    List<List<Integer>> list = new ArrayList<>();
    Arrays.sort(nums);
    backtrack(list, new ArrayList<>(), nums, target, 0);
    return list;
    
}

private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){
    if(remain < 0) return;
    else if(remain == 0) list.add(new ArrayList<>(tempList));
    else{
        for(int i = start; i < nums.length; i++){
            if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates
            tempList.add(nums[i]);
            backtrack(list, tempList, nums, remain - nums[i], i + 1);
            tempList.remove(tempList.size() - 1); 
        }
    }
}

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7 Palindrome Partitioning

public class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> res = new ArrayList<>();
        if (s.length() == 0) return res;
        dfs(s, 0, new ArrayList<String>(), res);
        return res;
    }
    public void dfs(String s, int index, ArrayList<String> item, List<List<String>> res){
        if (s.length() == index) {
            res.add(new ArrayList<>(item));
            return;
        }
        for (int i = index; i < s.length(); i++){
            if (isP(s, index, i)){
                item.add(s.substring(index, i + 1));
                dfs(s, i + 1, item, res);
                item.remove(item.size() - 1);
            }
        }
    }
    public boolean isP(String s, int l, int r){
        while (l < r) {
            if (s.charAt(l) != s.charAt(r)){
                return false;
            }
            l++;r--;
        }
        return true;
    }
}

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子集和排列

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