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subsets

2024-09-23 07:57:27   215人阅读

该篇总结了leetcode中 78. Subsets 和 90. Subsets II，主要算法思想是DFS

78. Subsets

Given a set of distinct integers, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]
public class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        
        if (nums == null) {
            return results;
        }
        
        if (nums.length == 0) {
            results.add(new ArrayList<Integer>());
            return results;
        }
        
        Arrays.sort(nums);
        helper(new ArrayList<Integer>(), nums, 0, results);
        return results;
    }
    
    
    // 递归三要素
    // 1. 递归的定义：在 Nums 中找到所有以 subset 开头的的集合，并放到 results
    //helper函数的参数： startIndex:从哪个下标开始将元素添加进子集；results:已经添加好的子集集合；subset：上次递归形成的子集，下次递归即要找以subset开头的集合
   
    private void helper(ArrayList<Integer> subset,
                        int[] nums,
                        int startIndex,
                        List<List<Integer>> results) {
        // 2. 递归的拆解
        // deep copy
        // results.add(subset);
        results.add(new ArrayList<Integer>(subset));
        
        for (int i = startIndex; i < nums.length; i++) {
            // [1] -> [1,2]
            subset.add(nums[i]);
            // 寻找所有以 [1,2] 开头的集合，并扔到 results
            helper(subset, nums, i + 1, results);
            // [1,2] -> [1]  注意回溯
            subset.remove(subset.size() - 1);
        }
        
    }
    
}

 

90. Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]
public class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        
        Arrays.sort(nums);
        
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        
        if(nums == null){
            return results;
        }
        
        if(nums.length == 0){
            results.add(new ArrayList<Integer>());
            return results;
        }
        
        helper(results, new ArrayList<Integer>(), nums, 0);
        
        return results;
    }
    
    private void helper(List<List<Integer>> results, ArrayList<Integer> list, int[] nums, int startIndex){
        
        results.add(new ArrayList<Integer>(list));
        
        for(int i = startIndex; i < nums.length; i++) {
            //与78.subsets比只增加了这一句
            if(i != startIndex && nums[i] == nums[i-1]){
                continue;
            }

            list.add(nums[i]);
            helper(results, list, nums, i+1);
            list.remove(list.size() - 1);
        }
    }
}

 

subsets


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