/**

* Difficulty:

* Medium

*

* Desc:

* Given a collection of integers that might contain duplicates, nums,

* return all possible subsets (the power set).

*

* Note:

* The solution set must not contain duplicate subsets.

*

* Example:

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

* [

* [2],

* [1],

* [1,2,2],

* [2,2],

* [1,2],

* []

* ]

*

* 和 No79. Subsets 差不多，但是数组中会有重复元素。

* 先排序，然后当元素重复出现时，直接利用上一次的结果，在上一次产生的子集中分布加入当前元素，生成新的子集

*/

/**

* @param {number[]} nums

* @return {number[][]}

*/

const subsetsWithDup = (nums) => {

nums.sort((a, b) => a - b);

const results = [[]];

let preArr = [];

for (let i = 0; i < nums.length; i += 1) {

let arr;

if (i > 0 && nums[i] === nums[i - 1]) {

arr = preArr.map((item) => {

return [

...item,

nums[i]

];

});

} else {

arr = results.map((item) => {

return [

...item,

nums[i]

];

});

}

results.push(...arr);

preArr = arr;

}

return results;

};