47 Permutations II
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,[1,1,2]
have the following unique permutations:
[
[1,1,2],
[1,2,1],
[2,1,1]
]
分析及代码
和46差不多,但是如果两个连续的数是一样的,并且前一个没有使用的话,后一个也没有必要尝试了,避免重复
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
if(nums.length == 0) {
return res;
}
Arrays.sort(nums);
boolean[] used = new boolean[nums.length];
generate(res, new ArrayList<>(), nums, used);
return res;
}
private void generate(List<List<Integer>> res, List<Integer> item, int[] nums, boolean[] used) {
if(item.size() == nums.length) {
res.add(new ArrayList<>(item));
return;
}
for(int i = 0; i < nums.length; i++) {
if(i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) {
continue;
}
if(!used[i]) {
used[i] = true;
item.add(nums[i]);
generate(res, item, nums, used);
used[i] = false;
item.remove(item.size() - 1);
}
}
}
O(n!)