[leetcode]#78 Subsets

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

Note:

Elements in a subset must be in non-descending order.

The solution set must not contain duplicate subsets.

For example,

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

[

[3],

[1],

[2],

[1,2,3],

[1,3],

[2,3],

[1,2],

[]

]

列出一个集合的所有子集

解题思路：1、看到这种生成多组数据的，而且富有规律性的，想到用递归生成。

2、将原集合的元素标号为0，1，2…n

3、递归调用后,每次将生成的集合压入容器a,并判断当前最大标号currentIndex是否为n,不为n,就继续循环添加元素currentIndex+1…n。

[code]#include <algorithmn>
class Solution {
public:
    void subset( vector<vector<int>>& a, vector<int>& nums, vector<int> b, int currentIndex) {
        sort(b.begin(), b.end());
        a.push_back(b);
        if (currentIndex+1 == nums.size())
            return;
        for (int i = currentIndex+1; i < nums.size(); i++) {
            vector<int> c = b;
            c.push_back(nums[i]);
            subset(a, nums, c, i);
        }
    }
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> a;
        vector<int> b;
        subset(a, nums, b, -1);
        return a;
    }
};