topic
地址：https://leetcode.com/problems/permutations-ii/
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
Example:
Input: [1,1,2]
Output:
[
[1,1,2],
[1,2,1],
[2,1,1]
]
Backtracking DFS achieve
Ideas:
- To the array nums sort;
- With Boolean type array used whether each location have been used record.
- Focus condition
if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]), if not the first position, and a position to keep up the value of the same, and have not used, do not explain the position of the head. For counter-examples[1, 1, 2], without this condition, two results can occur. This condition also[2, 1, 1]to de-emphasis, such as 1 second at position 1, because of the conditions!used[i - 1] == true, the success of de-emphasis. - Normally choose,
list.remove(list.size() - 1)the statement said they did not choose,used[i] = falseand the reset state.
package backtracking;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
// https://leetcode.com/problems/permutations-ii/
public class PermutationsII {
public static void main(String[] args) {
PermutationsII obj = new PermutationsII();
int[] input = {1,1,2};
List<List<Integer>> resultList = obj.permuteUnique(input);
System.out.println(Arrays.toString(resultList.toArray()));
}
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> resultList = new ArrayList<List<Integer>>();
Arrays.sort(nums);
boolean[] used = new boolean[nums.length];
if (nums == null || nums.length == 0) {
return resultList;
}
dfs(nums, resultList, new ArrayList<Integer>(), used);
return resultList;
}
private void dfs(int[] nums, List<List<Integer>> resultList, List<Integer> list, boolean[] used) {
if (list.size() == nums.length) {
resultList.add(new ArrayList<Integer>(list));
return;
}
for (int i = 0; i < nums.length; i++) {
if (used[i]) {
continue;
}
if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) {
continue;
}
used[i] = true;
list.add(nums[i]);
dfs(nums, resultList, list, used);
used[i] = false;
list.remove(list.size() - 1);
}
}
}
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