Lintcode 17 Subsets

Subsets 17

Question

Given a set of distinct integers, return all possible subsets.ExampleIf S = [1,2,3], a solution is:[ [3],[1],[2],[1,2,3],[1,3],[2,3],[1,2],[] ]ChallengeCan you do it in both recursively and iteratively?

Solution

Recursion类似Permutations，但是多传入一个参数pos用于记录位置，能加入的元素只能在pos和其之后的元素中寻找。同时，subsets长度可以任意，因此list在加入result时不用像permutation那样判断是否长度等于原数组长度。
Non-Recursion可以用位运算来实现。输入数组长度为n，则共有2^n个subset。从0到n－1位上，若用1表示取该位数，0表示不取该位数，则可以将这2^n种subset表示成0-2^n-1之间的二进制数。以[1,2,3]为例：

0 -> 000 -> []

1 -> 001 -> [1]

2 -> 010 -> [2]

..

7 -> 111 -> [1,2,3]

每次取一种subset的排列，看其在0-n－1位上的哪一位不为0，就将那一位上的数加入list。看第j位是否为0，可以和1 << j （j = 0, 1, ..., n-1）取&运算，若结果不为0，则第j位不为0。代码如下：Recursion:

class Solution {
/**
* @param S: A set of numbers.
* @return: A list of lists. All valid subsets.
*/
public ArrayList<ArrayList<Integer>> subsets(int[] nums) {
// write your code here
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> list = new ArrayList<Integer>();

if(nums == null || nums.length == 0){
return result;
}
Arrays.sort(nums);
helper(result, list, nums, 0);
return result;
}

public void helper(ArrayList<ArrayList<Integer>> result, ArrayList<Integer> list, int[] nums, int pos){
result.add(new ArrayList<Integer>(list));
//用pos记录当前位置，所有之后加入的元素只能在pos之后的元素中寻找
for(int i = pos; i < nums.length; i++){
list.add(nums[i]);
helper(result, list, nums, i+1);
list.remove(list.size()-1);
}
}
}

Non-Recursion:

class Solution {
/**
* @param S: A set of numbers.
* @return: A list of lists. All valid subsets.
*/
public ArrayList<ArrayList<Integer>> subsets(int[] nums) {
// write your code here
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
if(nums == null || nums.length == 0){
return result;
}

int n = nums.length;
Arrays.sort(nums);
for(int i = 0; i < (1 << n); i++){
ArrayList<Integer> list = new ArrayList<Integer>();
for(int j = 0; j < n; j++){
if((i & (1 << j)) != 0){
list.add(nums[j]);
}
}
result.add(list);
}

return result;
}
}

相关链接：https://www.kancloud.cn/kancloud/data-structure-and-algorithm-notes/73049 https://www.jiuzhang.com/solution/subsets/#tag-other