subsets-ii（需要思考，包括了子数组的求法）

还是有一定难度的。

基本方法，就是用队列，然后不断累加新的数。这是为了不重复而量身定制的。

如果运行重复，是有更简单清晰的方法，就是每次增加考虑一个数字，然后加到本来每一个结果的后面。如下：

public class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        res.add(new ArrayList<>()); 

        for (int num: nums) {
            List<List<Integer>> resDup = new ArrayList<>(res);
            for (List<Integer> list:resDup) {
                List<Integer> tmpList = new ArrayList<>(list);
                list.add(num);
                res.add(tmpList);
            }
        }
        return res;
    }
}

针对这道题目的解法：

https://leetcode.com/problems/subsets-ii/

// 好像跟之前也用的类似的方法

package com.company;

import java.util.*;

class Solution {
    class Pos {
        int pos;
        int len;
        Pos(int pos, int len) {
            this.pos = pos;
            this.len = len;
        }
    }

    public List<List<Integer>> subsetsWithDup(int[] nums) {
        Arrays.sort(nums);
        Queue<List<Integer>> qe = new ArrayDeque();
        Map<Integer, Pos> mp = new HashMap();
        List<List<Integer>> ret = new ArrayList<>();

        int len = 0;
        for (int i=0; i<nums.length; i++) {
            if (i == 0 || nums[i] == nums[i-1]) {
                len++;
            }
            else {
                Pos pos = new Pos(i-len, len);
                mp.put(nums[i-1], pos);
                len = 1;
            }
        }

        Pos pos = new Pos(nums.length-len, len);
        mp.put(nums[nums.length-1], pos);

        List<Integer> lst = new ArrayList();
        qe.offer(lst);
        ret.add(lst);

        while (!qe.isEmpty()) {
            List<Integer> tmpLst = qe.poll();
            boolean empty = true;
            int lastInt = 0;
            int curSize = -1;
            int curTail = nums[0];

            if (!tmpLst.isEmpty()) {
                empty = false;
                lastInt = tmpLst.get(tmpLst.size() - 1);
                curSize = tmpLst.size() - tmpLst.indexOf(lastInt);
                curTail = lastInt;
            }

            while (true) {
                Pos tmpPos = mp.get(curTail);
                if (empty || curTail > lastInt || tmpPos.len > curSize) {
                    List<Integer> inputLst = new ArrayList<>(tmpLst);
                    inputLst.add(curTail);
                    qe.offer(inputLst);
                    ret.add(inputLst);
                }
                if (tmpPos.pos + tmpPos.len >= nums.length) {
                    break;
                }
                curTail = nums[tmpPos.pos + tmpPos.len];
            }

        }
        return ret;
    }
}

public class Main {

    public static void main(String[] args) {
        System.out.println("Hello!");
        Solution solution = new Solution();

        int[] nums = {1,1,2,2,2};
        List<List<Integer>> ret = solution.subsetsWithDup(nums);
        System.out.printf("Get ret: %d\n", ret.size());
        Iterator<List<Integer>> iter = ret.iterator();
        while (iter.hasNext()) {
            Iterator itemItr = iter.next().iterator();
            while (itemItr.hasNext()) {
                System.out.printf("%d,", itemItr.next());
            }
            System.out.println();
        }

        System.out.println();

    }
}

// 这是之前的方法

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<vector<int>> result;
        sort(nums.begin(), nums.end());

        vector<int> tmp;
        result.push_back(tmp);

        int vlen;
        int is_dup = 0;
        vector<int> newtmp;

        int nlen = nums.size();
        for (int i = 0; i < nlen; i++) {
            if (i > 0 && nums[i] == nums[i-1]) {
                is_dup++;
            }
            else {
                is_dup = 0;
            }

            vlen = result.size();
            for (int j = 0; j < vlen; j++) {
                tmp = result[j];
                if (is_dup > 0 && \
                    (tmp.size() < is_dup || tmp[tmp.size()-is_dup] != nums[i])) {
                    // ignore dup
                    continue;
                }
                newtmp.resize(tmp.size());
                copy(tmp.begin(), tmp.end(), newtmp.begin());
                newtmp.push_back(nums[i]);
                result.push_back(newtmp);
            }
        }

        return result;

    }
};