[LeetCode] Subsets 子集合

Given a set of distinct integers, S, 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 S = 

[1,2,3]

[code][
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]

 解题思路：

这道题目拿到手第一想法就是利用DFS，与以往不同的是子集的大小是从0到3，这个点需要额外注意，本初就是在主函数中构造一个for循环来解决这个问题，剩下的就是整张的dfs算法。当然我在此处天添加了visited 和 start，就我个人的经验来说，我认为i这种集合顺序性存在的题目，怎么说呢 就是 12345567  用第一个再用第二个再用第三个的 用start明显好一些。

[code]class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> subset = new ArrayList<>();
List<Integer> subsetList = new ArrayList<>();
boolean[] visited = new boolean[nums.length];
for(int i=0; i <= nums.length; i++)
help(subset, subsetList, visited, 0, i, nums);
return subset;
}
private void help(List<List<Integer>> subset, List<Integer> subsetList,
boolean[] visited, int start, final int size, final int[] nums){
if(subsetList.size() == size){
subset.add(new ArrayList<>(subsetList));
return;
}

for(int i = start; i < nums.length; i++){
subsetList.add(nums[i]);
help(subset, subsetList, visited, i + 1, size, nums);
subsetList.remove(subsetList.size() - 1);
}
}
}

[code]class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> list = new ArrayList<>();
result.add(new ArrayList());
boolean[] visited = new boolean[nums.length];
help(list, result, nums, 0, visited);
return result;
}

private void help(List<Integer> list, List<List<Integer>> result, int[] nums, int start,
boolean[] visited){
for(int i = start; i < nums.length; i++){
visited[i] = true;
list.add(nums[i]);
if(list.size() <= nums.length)
result.add(new ArrayList(list));
help(list, result, nums, i+1, visited);
list.remove(list.size() - 1);
}

}
}