78. 子集

输入: nums = [1,2,3]
输出:
[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

/*
 * @lc app=leetcode id=78 lang=javascript
 *
 * [78] Subsets
 *
 * https://leetcode.com/problems/subsets/description/
 *
 * algorithms
 * Medium (51.19%)
 * Total Accepted:    351.6K
 * Total Submissions: 674.8K
 * Testcase Example:  '[1,2,3]'
 *
 * Given a set of distinct integers, nums, return all possible subsets (the
 * power set).
 *
 * Note: The solution set must not contain duplicate subsets.
 *
 * Example:
 *
 *
 * Input: nums = [1,2,3]
 * Output:
 * [
 * ⁠ [3],
 * [1],
 * [2],
 * [1,2,3],
 * [1,3],
 * [2,3],
 * [1,2],
 * []
 * ]
 *
 */
function backtrack(list, tempList, nums, start) {
    list.push([...tempList]);
    for(let i = start; i < nums.length; i++) {
        tempList.push(nums[i]);
        backtrack(list, tempList, nums, i + 1);
        tempList.pop();
    }
}
/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var subsets = function(nums) {
    const list = [];
    backtrack(list, [], nums, 0);
    return list;
};

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        auto ret = vector<vector<int>>();
        auto tmp = vector<int>();
        backtrack(ret, tmp, nums, 0);
        return ret;
    }

    void backtrack(vector<vector<int>>& list, vector<int>& tempList, vector<int>& nums, int start) {
        list.push_back(tempList);
        for (auto i = start; i < nums.size(); ++i) {
            tempList.push_back(nums[i]);
            backtrack(list, tempList, nums, i + 1);
            tempList.pop_back();
        }
    }
};