import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
/**
* Source : https://oj.leetcode.com/problems/permutations-ii/
*
* Created by lverpeng on 2017/7/17.
*
* 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], and [2,1,1].
*
*/
public class Permutation2 {
/**
* 找出数组元素可以组成的所有排列
*
* 递归求出每一个元素开头的组合
*
*
* @param arr
* @return
*/
public void permute (int[] arr, int start, int end, List<int[]> result) {
if (start >= end-1) {int[] newArr = new int[arr.length];
System.arraycopy(arr,0, newArr, 0, arr.length);
result.add(newArr);
return ;
}
for (int i = start; i < end; i++) {
if (i < end - 1 && arr[i] == arr[i + 1]) {
// 如果是相同的元素,则跳过当前元素
continue;
}
swap(arr, start, i);
permute(arr, start + 1, end, result);
swap(arr, i, start);
}
}
private void swap (int[] arr, int i, int j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public static void printList (List<int[]> list) {
for (int i = 0; i < list.size(); i++) {
System.out.println(Arrays.toString(list.get(i)));
}
}
public static void main(String[] args) {
Permutation2 permutation2 = new Permutation2();
int[] arr = new int[]{1,2,1};
Arrays.sort(arr);
int[] arr1 = new int[]{1,3,2};
Arrays.sort(arr1);
List<int[]> result = new ArrayList<int[]>();
permutation2.permute(arr, 0, arr.length, result);
printList(result);
System.out.println();
result = new ArrayList<int[]>();
permutation2.permute(arr1, 0, arr1.length, result);
printList(result);
}
}