LeetCode78
2017-01-05 23:46
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这道题比较简单~一次就AC了~
但是我觉得求求集合子集的算法是一个很重要的子算法,还是在这里记录一下自己的思路。
Given a set of distinct integers, nums, return all possible subsets.
Note: The solution set must not contain duplicate subsets.
For example,
If nums =
is:
上边是题目要求
我的思路是类似二叉树穷举,在每层分叉时进行两次递归,一次加入待选元素,一次不加入。
public class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> retlist = new ArrayList<List<Integer>>();
List<Integer> cur = new ArrayList<Integer>();
tree(retlist,cur,0,nums);
return retlist;
}
private void tree (List<List<Integer>> retlist,List<Integer> cur,int level,int[] nums){
List<Integer> curnew = new ArrayList<Integer>();
curnew.addAll(cur);
if (level == nums.length){
retlist.add(curnew);
return;
}
tree(retlist,curnew,level+1,nums);
curnew.add(nums[level]);
tree(retlist,curnew,level+1,nums);
}
}
但是我觉得求求集合子集的算法是一个很重要的子算法,还是在这里记录一下自己的思路。
Given a set of distinct integers, nums, return all possible subsets.
Note: The solution set must not contain duplicate subsets.
For example,
If nums =
[1,2,3], a solution
is:
上边是题目要求
我的思路是类似二叉树穷举,在每层分叉时进行两次递归,一次加入待选元素,一次不加入。
public class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> retlist = new ArrayList<List<Integer>>();
List<Integer> cur = new ArrayList<Integer>();
tree(retlist,cur,0,nums);
return retlist;
}
private void tree (List<List<Integer>> retlist,List<Integer> cur,int level,int[] nums){
List<Integer> curnew = new ArrayList<Integer>();
curnew.addAll(cur);
if (level == nums.length){
retlist.add(curnew);
return;
}
tree(retlist,curnew,level+1,nums);
curnew.add(nums[level]);
tree(retlist,curnew,level+1,nums);
}
}
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