/**

* Difficulty:

* Medium

*

* Desc:

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

*

* Example:

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

* [

* [3],

* [1],

* [2],

* [1,2,3],

* [1,3],

* [2,3],

* [1,2],

* []

* ]

*

* Note:

* The solution set must not contain duplicate subsets.

*

* 从数组中求出所有没有重复元素的组合

* 跟上一题 No77.Combinations 一样的思路，只是稍微复杂了一点

*/

/**

* @param {number[]} nums

* @return {number[][]}

*/

var subsets = function(nums) {

var results = [[]];

var max = nums.length;

var chooseNum = function(result, min, layer, maxLayer) {

for (var i = min + 1; i < max; i += 1) {

result.push(nums[i]);

if (layer === maxLayer) {

results.push(result.slice(0));

} else {

chooseNum(result, i, layer + 1, maxLayer);

}

result.pop();

}

}

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

var num = nums[i];

results.push([num]);

var length = 2;

while(length <= nums.length - i) {

chooseNum([num], i, 2, length);

length += 1;

}

}

return results;

};

/**

* Solution2:

* 已知数组中没有重复元素，所以我们不用考虑过滤问题

* 遍历数组，对于每一个元素，都把它加入到之前生成的子集中，返回一组新的子集，这个子集就是该元素所能够生成的子集

* 然后把这组子集加入到要返回的结果中去

*/

const subsets = (nums) => {

const results = [[]];

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

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

return [

...item,

nums[i]

];

});

results.push(...arr);

}

return results;

};