[leetcode]47. Permutations II@Java解题报告

https://leetcode.com/problems/permutations-ii/description/

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]
]

package go.jacob.day819;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
* 47. Permutations II
*
* @author Jacob
* 思路：与46. Permutations解法类似。区别是，该题需要单独建立一个boolean数组
* 用来记录nums中的元素是否被使用(是否已经加入list集合)
*/
public class Demo2 {
List<List<Integer>> res;

public List<List<Integer>> permuteUnique(int[] nums) {
res = new ArrayList<List<Integer>>();
if (nums == null || nums.length < 1)
return res;
Arrays.sort(nums);
ArrayList<Integer> list = new ArrayList<Integer>();
boolean[] used = new boolean[nums.length];
solve(list, nums, used);

return res;
}

private void solve(ArrayList<Integer> list, int[] nums, boolean[] used) {
if (list.size() == nums.length) {
res.add(new ArrayList<Integer>(list));
return;
}

for (int i = 0; i < nums.length; i++) {
if (used[i])
continue;
/*
* 只需要判断i和i-1(而不需要判断i与i-2...) 相同的元素一定是相邻的。
* 如果i-2，i-1，i相同，那么在上一轮循环就已经判断了i-1,i，本轮循环不需要重复判断
*/
if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1])
continue;
list.add(nums[i]);
used[i]=true;
solve(list, nums, used);
used[i]=false;
list.remove(list.size()-1);
}

}
}