Question
Formatted question description: https://leetcode.ca/all/40.html

```
40. 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, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
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]
```

Algorithm
Based on Combination_Sum_I, add sorting to move duplicated numbers together and skip.

Code
Java

import java.util.ArrayList ;
import java.util.Arrays ;
import java.util.List ;
public class Combination_Sum_II {
class Solution {
List < List < Integer >> result = new ArrayList <>();
public List < List < Integer >> combinationSum2 ( int [] candidates , int target ) {
if ( candidates == null || candidates . length == 0 ) {
return result ;
}
// nlogn
Arrays . sort ( candidates );
dfs ( candidates , target , 0 , new ArrayList < Integer >());
return result ;
}
private void dfs ( int [] candidates , int target , int start , ArrayList < Integer > onePath ) {
if ( target < 0 ) { // target<0, assumption is all positive
return ;
}
if ( target == 0 ) {
result . add ( new ArrayList <>( onePath ));
// continue, and possible 1,-1 are next ones
}
for ( int i = start ; i < candidates . length ; i ++) {
if ( i > start && candidates [ i ] == candidates [ i - 1 ]) continue ;
onePath . add ( candidates [ i ]);
dfs ( candidates , target - candidates [ i ], i + 1 , onePath ); // i+1, to use only once
onePath . remove ( onePath . size () - 1 );
}
}
}
// https://leetcode.com/problems/combination-sum-ii/solution/
// official solution got "Time Limit Exceeded" error, funny but still a good idea solving this problem
class Solution_lc {
public List < List < Integer >> combinationSum2 ( int [] candidates , int target ) {
// container to hold the final combinations
List < List < Integer >> results = new ArrayList <>();
LinkedList < Integer > comb = new LinkedList <>();
HashMap < Integer , Integer > counter = new HashMap <>();
for ( int candidate : candidates ) {
if ( counter . containsKey ( candidate ))
counter . put ( candidate , counter . get ( candidate ) + 1 );
else
counter . put ( candidate , 1 );
}
// convert the counter table to a list of (num, count) tuples
List < int []> counterList = new ArrayList <>();
counter . forEach (( key , value ) -> {
counterList . add ( new int []{ key , value });
});
backtrack ( comb , target , 0 , counterList , results );
return results ;
}
private void backtrack ( LinkedList < Integer > comb ,
int remain , int curr ,
List < int []> counter ,
List < List < Integer >> results ) {
if ( remain == 0 ) {
// make a deep copy of the current combination.
results . add ( new ArrayList < Integer >( comb ));
return ;
}
for ( int nextCurr = curr ; nextCurr < counter . size (); ++ nextCurr ) {
int [] entry = counter . get ( nextCurr );
Integer candidate = entry [ 0 ], freq = entry [ 1 ];
if ( freq <= 0 )
continue ;
// add a new element to the current combination
comb . addLast ( candidate );
counter . set ( nextCurr , new int []{ candidate , freq - 1 });
// continue the exploration with the updated combination
backtrack ( comb , remain - candidate , nextCurr , counter , results );
// backtrack the changes, so that we can try another candidate
counter . set ( nextCurr , new int []{ candidate , freq });
comb . removeLast ();
}
}
}
}