78. Subsets
题目:
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]
, a solution is:
[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]
链接: http://leetcode.com/problems/subsets/
题解:
求数组子数组。先把数组排序,之后就可以使用DFS,维护一个递增的position,递归后要backtracking。
Time Complexity - O(n * 2n), Space Complexity - O(n)
public class Solution {
public ArrayList<ArrayList<Integer>> subsets(int[] S) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
if(S == null || S.length == 0)
return result;
ArrayList<Integer> list = new ArrayList<Integer>();
Arrays.sort(S);
helper(result, list, S, 0);
return result;
} private void helper(ArrayList<ArrayList<Integer>> result, ArrayList<Integer> list, int[] S, int pos){
result.add(new ArrayList<Integer>(list)); for(int i = pos; i < S.length; i ++){
list.add(S[i]);
helper(result, list, S, ++pos);
list.remove(list.size() - 1);
}
}
}
Updates:
public class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
if(nums == null || nums.length == 0)
return res;
Arrays.sort(nums);
ArrayList<Integer> list = new ArrayList<>();
dfs(res, list, nums, 0);
return res;
} private void dfs(List<List<Integer>> res, ArrayList<Integer> list, int[] nums, int pos) {
res.add(new ArrayList<Integer>(list)); for(int i = pos; i < nums.length; i++) {
list.add(nums[i]);
dfs(res, list, nums, ++pos);
list.remove(list.size() - 1);
}
}
}
Update:
为什么以前总写成++pos? i + 1就可以了
public class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
if(nums == null || nums.length == 0)
return res;
Arrays.sort(nums);
ArrayList<Integer> list = new ArrayList<>();
dfs(res, list, nums, 0);
return res;
} private void dfs(List<List<Integer>> res, ArrayList<Integer> list, int[] nums, int pos) {
res.add(new ArrayList<Integer>(list)); for(int i = pos; i < nums.length; i++) {
list.add(nums[i]);
dfs(res, list, nums, i + 1);
list.remove(list.size() - 1);
}
}
}
二刷:
发现自己以前不懂装懂糊弄过去了好多题...我勒个去。
这道题目我们也是使用跟上一题combination类似的方法。
- 这里我们根据题意首先要对数组排个序
- 构造一个辅助函数getSubsets来进行DFS和backtracking, 同时这个辅助函数还要有一个pos来控制遍历的位置,我们先pass 0 进去。
- 每次进入getSubsets我们都直接往结果集中加入一个当前List的新副本
- 接下来从pos开始遍历整个数组,每次进入新一层dfs的时候pass 新的pos = i + 1,这样就能保证结果中的顺序是从小到大
Java:
Time Complexity - O(n!), Space Complexity (n2)
public class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
if (nums == null || nums.length == 0) {
return res;
}
Arrays.sort(nums);
List<Integer> list = new ArrayList<>();
getSubsets(res, list, nums, 0);
return res;
} private void getSubsets(List<List<Integer>> res, List<Integer> list, int[] nums, int pos) {
res.add(new ArrayList<Integer>(list));
for (int i = pos; i < nums.length; i++) {
list.add(nums[i]);
getSubsets(res, list, nums, i + 1);
list.remove(list.size() - 1);
}
}
}
三刷:
下次还需要研究Bit Manipulation 以及 iterative的写法。
Java:
Time Complexity - O(n * 2n), Space Complexity (2n)
class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
if (nums == null || nums.length == 0) return res;
subsets(res, nums, new ArrayList<Integer>(), 0);
return res;
} private void subsets(List<List<Integer>> res, int[] nums, List<Integer> list, int idx) {
res.add(new ArrayList<>(list)); for (int i = idx; i < nums.length; i++) {
list.add(nums[i]);
subsets(res, nums, list, i + 1);
list.remove(list.size() - 1);
}
}
}
测试:
Reference:
https://leetcode.com/discuss/72498/simple-iteration-no-recursion-no-twiddling-explanation
https://leetcode.com/discuss/25696/simple-java-solution-with-for-each-loops
https://leetcode.com/discuss/29631/java-subsets-solution
https://leetcode.com/discuss/46668/recursive-iterative-manipulation-solutions-explanations
https://leetcode.com/discuss/9213/my-solution-using-bit-manipulation
http://www.cnblogs.com/springfor/p/3879830.html
http://www.cnblogs.com/zhuli19901106/p/3492515.html
http://www.1point3acres.com/bbs/thread-117602-1-1.html
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