Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a₁, a₂, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• The solution set must not contain duplicate combinations.

 

For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6]

 

 1 public class Solution {

 2     ArrayList<ArrayList<Integer>> result = null;

 3     public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {

 4         // Note: The Solution object is instantiated only once and is reused by each test case.

 5         Arrays.sort(num);

 6         result = new ArrayList<ArrayList<Integer>>();

 7         getSolution(new ArrayList<Integer>(), target, num, 0);

 8         return result;

 9     }

10     public void getSolution(ArrayList<Integer> row, int target, int[] num, int pos){

11         int i = pos;

12         int per = -1;

13         if(target == 0){

14             result.add(row);

15         }

16         while(i < num.length && num[i] <= target){

17             if(per != num[i]){

18                 per = num[i];

19                 row.add(num[i]);

20                 target -= num[i];

21                 int tmp = num[i];
23                 getSolution(new ArrayList<Integer>(row), target, num, i + 1);

24                 row.remove(row.size() - 1);
26                 target += num[i];

27             }

28             i ++;

29         }

30     }

31 }

 第三遍：

test case:

Input:	[1,1], 1
Output:	[[1],[1]]
Expected:	[[1]]

结果相同如何处理。

 1 public class Solution {

 2     public List<List<Integer>> combinationSum2(int[] candidates, int target) {

 3         Arrays.sort(candidates);

 4         List<List<Integer>> result = new ArrayList<List<Integer>>();

 5         findElement(candidates, 0, candidates.length, result, new ArrayList<Integer>(), target);

 6         return result;

 7     }

 8     

 9     public void findElement(int[] arr, int start, int end, List<List<Integer>> result, ArrayList<Integer> row, int tar){

10         if(tar == 0) result.add(row);

11         for(int i = start; i < end; i ++){

12             if((i == start || arr[i] != arr[i - 1]) && arr[i] <= tar){

13                 ArrayList<Integer> rown = new ArrayList<Integer> (row);

14                 rown.add(arr[i]);

15                 findElement(arr, i + 1, end, result, rown, tar - arr[i]);

16             }

17         }

18     }

19 }

(i == start || arr[i] != arr[i - 1]) 保证了在同一层里不会选取一样的数字。