leetcode : subsetsII


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) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if(nums == null || nums.length == 0) {
            return result;
        }
        List<Integer> list = new ArrayList<Integer>();
        Arrays.sort(nums);
        backtrack(result, list, nums, 0);
        return result;
    }

    public void backtrack(List<List<Integer>> result, List<Integer> list, int[] nums, int position) {
        if(list.size() > nums.length) {
            return;
        }

        result.add(new ArrayList<Integer>(list));

        for(int i = position; i < nums.length; i++) {
            if(i > position && nums[i] == nums[i - 1]) {
                continue;
            }
            list.add(nums[i]);
            backtrack(result, list, nums, i + 1);
            list.remove(list.size() - 1);
        }
    }

}
时间: 2024-12-28 12:35:04

leetcode : subsetsII的相关文章

从n个元素中选择k个的所有组合(包含重复元素)

LeetCode:Combinations这篇博客中给出了不包含重复元素求组合的5种解法.我们在这些解法的基础上修改以支持包含重复元素的情况.对于这种情况,首先肯定要对数组排序,以下不再强调 修改算法1:按照求包含重复元素集合子集的方法LeetCode:Subsets II算法1的解释,我们知道:若当前处理的元素如果在前面出现过m次,那么只有当前组合中包含m个该元素时,才把当前元素加入组合 ? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Leetcode: Subsets &amp; SubsetsII

Subsets Description: Given a set of distinct integers, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example,If S = [1,2,3], a solution is: [ [3],

[leetcode]Subsets II @ Python

原题地址:https://oj.leetcode.com/problems/subsets-ii/ 题意: Given a collection of integers that might contain duplicates, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicat

LeetCode题目总结分类

注:此分类仅供大概参考,没有精雕细琢.有不同意见欢迎评论~ 利用堆栈:http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode.com/problems/longest-valid-parentheses/ (也可以用一维数组,贪心)http://oj.leetcode.com/problems/valid-parentheses/http://oj.leetcode.com/probl

[LeetCode]题解(python):090 Subsets II

题目来源 https://leetcode.com/problems/subsets-ii/ Given a collection of integers that might contain duplicates, nums, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate su

[LeetCode] 349 Intersection of Two Arrays &amp; 350 Intersection of Two Arrays II

这两道题都是求两个数组之间的重复元素,因此把它们放在一起. 原题地址: 349 Intersection of Two Arrays :https://leetcode.com/problems/intersection-of-two-arrays/description/ 350 Intersection of Two Arrays II:https://leetcode.com/problems/intersection-of-two-arrays-ii/description/ 题目&解法

LeetCode 442. Find All Duplicates in an Array (在数组中找到所有的重复项)

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements that appear twice in this array. Could you do it without extra space and in O(n) runtime? Example: Input: [4,3,2,7,

LeetCode OJ - Sum Root to Leaf Numbers

这道题也很简单,只要把二叉树按照宽度优先的策略遍历一遍,就可以解决问题,采用递归方法越是简单. 下面是AC代码: 1 /** 2 * Sum Root to Leaf Numbers 3 * 采用递归的方法,宽度遍历 4 */ 5 int result=0; 6 public int sumNumbers(TreeNode root){ 7 8 bFSearch(root,0); 9 return result; 10 } 11 private void bFSearch(TreeNode ro

LeetCode OJ - Longest Consecutive Sequence

这道题中要求时间复杂度为O(n),首先我们可以知道的是,如果先对数组排序再计算其最长连续序列的时间复杂度是O(nlogn),所以不能用排序的方法.我一开始想是不是应该用动态规划来解,发现其并不符合动态规划的特征.最后采用类似于LRU_Cache中出现的数据结构(集快速查询和顺序遍历两大优点于一身)来解决问题.具体来说其数据结构是HashMap<Integer,LNode>,key是数组中的元素,所有连续的元素可以通过LNode的next指针相连起来. 总体思路是,顺序遍历输入的数组元素,对每个