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0090. Subsets II
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markdown/0090. Subsets II.md

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### [90\. Subsets II](https://leetcode.com/problems/subsets-ii/)
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Difficulty: **Medium**
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Given a collection of integers that might contain duplicates, **_nums_**, return all possible subsets (the power set).
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**Note:** The solution set must not contain duplicate subsets.
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**Example:**
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```
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Input: [1,2,2]
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Output:
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[
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[2],
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[1],
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[1,2,2],
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[2,2],
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[1,2],
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[]
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]
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```
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#### Solution
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Language: **Java**
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```java
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class Solution {
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   public List<List<Integer>> subsetsWithDup(int[] nums) {
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       List<List<Integer>> result = new ArrayList<>();
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       List<Integer> elem = new ArrayList<>();
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       result.add(elem);
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       Arrays.sort(nums);
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       for (int i = 0; i < nums.length; i++) {
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           int dupCount = 0;
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           while (((i + 1) < nums.length) && nums[i + 1] == nums[i]) {
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               dupCount++;
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               i++;
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          }
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           int size = result.size();
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           for (int j = 0; j < size; j++) {
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               elem = new ArrayList<>(result.get(j));
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               for (int k = 0; k <= dupCount; k++) {
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                   elem.add(nums[i]);
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                   result.add(new ArrayList<>(elem)); // 注意这里是要放在 for 循环里面的
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              }
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          }
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      }
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       return result;
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  }
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}
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```
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![pic](https://raw.githubusercontent.com/PicGoBed/PicBed/master/2019-08-03-1F2FZ6.jpg)

src/main/java/leetcode/_90_/Main.java

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package leetcode._90_;
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/**
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* Created by zhangbo54 on 2019-03-04.
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*/
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public class Main {
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public static void main(String[] args) {
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Solution solution = new Solution();
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int[] nums = {1, 2, 2};
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System.out.println(solution.subsetsWithDup(nums));
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}
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}
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package leetcode._90_;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.List;
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class Solution {
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public List<List<Integer>> subsetsWithDup(int[] nums) {
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List<List<Integer>> result = new ArrayList<>();
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List<Integer> elem = new ArrayList<>();
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result.add(elem);
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Arrays.sort(nums);
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for (int i = 0; i < nums.length; i++) {
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int dupCount = 0;
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while (((i + 1) < nums.length) && nums[i + 1] == nums[i]) {
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dupCount++;
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i++;
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}
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int size = result.size();
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for (int j = 0; j < size; j++) {
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elem = new ArrayList<>(result.get(j));
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for (int k = 0; k <= dupCount; k++) {
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elem.add(nums[i]);
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result.add(new ArrayList<>(elem)); // 注意这里是要放在 for 循环里面的
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}
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}
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}
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return result;
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}
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}
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### [90\. Subsets II](https://leetcode.com/problems/subsets-ii/)
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Difficulty: **Medium**
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Given a collection of integers that might contain duplicates, **_nums_**, return all possible subsets (the power set).
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**Note:** The solution set must not contain duplicate subsets.
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**Example:**
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```
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Input: [1,2,2]
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Output:
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[
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[2],
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[1],
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[1,2,2],
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[2,2],
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[1,2],
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[]
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]
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```
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#### Solution
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Language: **Java**
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```java
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class Solution {
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   public List<List<Integer>> subsetsWithDup(int[] nums) {
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       List<List<Integer>> result = new ArrayList<>();
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       List<Integer> elem = new ArrayList<>();
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       result.add(elem);
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       Arrays.sort(nums);
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       for (int i = 0; i < nums.length; i++) {
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           int dupCount = 0;
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           while (((i + 1) < nums.length) && nums[i + 1] == nums[i]) {
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               dupCount++;
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               i++;
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          }
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           int size = result.size();
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           for (int j = 0; j < size; j++) {
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               elem = new ArrayList<>(result.get(j));
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               for (int k = 0; k <= dupCount; k++) {
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                   elem.add(nums[i]);
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                   result.add(new ArrayList<>(elem)); // 注意这里是要放在 for 循环里面的
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              }
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          }
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      }
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       return result;
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  }
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}
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```
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![pic](https://raw.githubusercontent.com/PicGoBed/PicBed/master/2019-08-03-1F2FZ6.jpg)

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