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]
, and [2,1,1]
.
和permutations 基本一致, 只是需要先把array 排序,然后保证在同一次循环中选取的数字不同即可。
1 public class Solution { 2 ArrayList<ArrayList<Integer>> result = null; 3 int len = 0; 4 public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) { 5 // Start typing your Java solution below 6 // DO NOT write main() function 7 Arrays.sort(num); 8 result = new ArrayList<ArrayList<Integer>>(); 9 if(num == null || num.length == 0) return result; 10 len = num.length; 11 boolean[] item = new boolean[len]; 12 get(item, num, new ArrayList<Integer>()); 13 return result; 14 } 15 public void get(boolean[] item, int[] num, ArrayList<Integer> row){ 16 if(row.size() == len){ 17 result.add(new ArrayList<Integer>(row)); 18 return; 19 } 20 for(int i = 0; i < len; i ++){ 21 if(!item[i]){ 22 item[i] = true; 23 row.add(num[i]); 24 get(item, num, row); 25 item[i] = false; 26 row.remove(row.size() - 1); 27 while(i + 1 < num.length && num[i + 1] == num[i]) 28 i ++; 29 } 30 } 31 } 32 }
第三遍: 不使用额外的boolean 数组:
1 public class Solution { 2 public List<List<Integer>> permuteUnique(int[] num) { 3 ArrayList<List<Integer>> result = new ArrayList<List<Integer>>(); 4 Arrays.sort(num); 5 getPermutation(num, num.length, new ArrayList<Integer>(), result); 6 return result; 7 } 8 public void getPermutation(int[] num, int len, ArrayList<Integer> row, ArrayList<List<Integer>> result){ 9 if(num == null || len == 0){ 10 result.add(row); 11 return; 12 } 13 int pre = Integer.MAX_VALUE; 14 for(int i = 0; i < num.length; i ++){ 15 if(num[i] != Integer.MAX_VALUE && num[i] != pre){ 16 pre = num[i]; 17 num[i] = Integer.MAX_VALUE; 18 ArrayList<Integer> nrow = new ArrayList<Integer>(row); 19 nrow.add(pre); 20 getPermutation(num, len - 1, nrow, result); 21 num[i] = pre; 22 } 23 } 24 } 25 }
第四遍:
1 public class Solution { 2 ArrayList<List<Integer>> result; 3 public List<List<Integer>> permuteUnique(int[] num) { 4 result = new ArrayList<List<Integer>>(); 5 if(num == null || num.length == 0) return result; 6 Arrays.sort(num); 7 findPermutation(num, new ArrayList<Integer>()); 8 return result; 9 } 10 11 private void findPermutation(int[] num, ArrayList<Integer> row){ 12 if(row.size() == num.length){ 13 result.add(row); 14 } 15 int pre = Integer.MIN_VALUE; 16 for(int i = 0; i < num.length; i ++){ 17 if(num[i] != Integer.MIN_VALUE && num[i] != pre){ 18 pre = num[i]; 19 row.add(num[i]); 20 num[i] = Integer.MIN_VALUE; 21 findPermutation(num, new ArrayList<Integer>(row)); 22 num[i] = pre; 23 row.remove(row.size() - 1); 24 } 25 } 26 } 27 }
posted on 2013-11-07 12:53 Step-BY-Step 阅读(168) 评论(0) 编辑 收藏 举报