Combination Sum II

Combination Sum II (leetcodelintcode)

Description
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.

Notice
- 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.

Example
Given candidate set [10,1,6,7,2,1,5] and target 8,
A solution set is:
[
  [1,7],
  [1,2,5],
  [2,6],
  [1,1,6]
]

解题思路

参考 Combination Sum 的解题思路，区别在于不能重复选取数组元素，所以在搜索时下一层递归的起始位置加一。

Java 实现

public
class
Solution
{

/**
     * 
@param
 num: Given the candidate numbers
     * 
@param
 target: Given the target number
     * 
@return
: All the combinations that sum to target
     */
public
 List
<
List
<
Integer
>
>
 combinationSum2(
int
[] num, 
int
 target) {
        List
<
List
<
Integer
>
>
 result = 
new
 ArrayList
<
List
<
Integer
>
>
();

if
 (num == 
null
 || num.length == 
0
 || target 
<
= 
0
) {

//result.add(new ArrayList
<
Integer
>
());
return
 result;
        }

        Arrays.sort(num);
        List
<
Integer
>
 path = 
new
 ArrayList
<
Integer
>
();
        dfs(result, path, num, target, 
0
);

return
 result;
    }


private
void
dfs
(List
<
List
<
Integer
>
>
 result, List
<
Integer
>
 path,

int
[] num, 
int
 target, 
int
 pos)
{

if
 (target == 
0
) {
            result.add(
new
 ArrayList
<
Integer
>
(path));
        }


for
 (
int
 i = pos; i 
<
 num.length; i++) {

if
 (num[i] 
>
 target) {

break
;
            }

if
 (i != pos 
&
&
 num[i] == num[i - 
1
]) {

continue
;
            }
            path.add(num[i]);
            dfs(result, path, num, target - num[i], i + 
1
);
            path.remove(path.size() - 
1
);
        }
    }
}