Subsets II ——LeetCode

Given a collection of integers that might contain duplicates, 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 =

[1,2,2]

, a solution is:

[
[2],
[1],
[1,2,2],
[2,2],
[1,2],
[]
]

题目大意：给定一个数组，返回所有的可能子集合，数组有重复元素，但是子集合不能重复。

解题思路：这题我首先sort一下，然后遍历，当前元素不是重复的元素就把res里的所有的list取出来加上当前的元素再添加到res里，是重复元素就把上一次加入res的此元素list取出并加上重复元素再放入res。

public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
if (nums == null || nums.length == 0) {
return res;
}
Arrays.sort(nums);
res.add(new ArrayList<Integer>());
int lastSize = 0, currSize;
for (int i = 0; i < nums.length; i++) {
int start = (i > 0 && nums[i] == nums[i - 1]) ? lastSize : 0;
currSize = res.size();
for (int j = start; j < currSize; j++) {
List<Integer> tmp = new ArrayList<>(res.get(j));
tmp.add(nums[i]);
res.add(tmp);
}
lastSize = currSize;
}
return res;
}