Combination Sum

Combination Sum (leetcodelintcode)

Description
Given a set of candidate numbers (C) and a target number (T), 
find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.

For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

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 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3]

解题思路

本题与 Subsets 不同之处在于同一个数组元素可以重复使用，以[2, 3, 6, 7]举例，第一个放入list中的是首元素2，下一次递归调用仍将2放入list中，直至list的和大于目标值，然后依次取出list队尾的2，开始将3放入，以此类推。

易错点

1. 原始数组中可能有重复元素，所以需要先将原始数组 排序 ，在函数中增加 去重处理 。

Java 实现

public
class
Solution
{

/**
     * 
@param
 candidates: A list of integers
     * 
@param
 target:An integer
     * 
@return
: A list of lists of integers
     */
public
 List
<
List
<
Integer
>
>
 combinationSum(
int
[] candidates, 
int
 target) {
        List
<
List
<
Integer
>
>
 result = 
new
 ArrayList
<
>
();

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

return
 result;
        }

        List
<
Integer
>
 list = 
new
 ArrayList
<
>
();
        Arrays.sort(candidates);
        search(candidates, target, 
0
, list, result);

return
 result;
    }


private
void
search
(
int
[] candidates,

int
 target,

int
 pos,
                        List
<
Integer
>
 list,
                        List
<
List
<
Integer
>
>
 result)
{


int
 count = sum(list);

if
 (count == target) {
            result.add(
new
 ArrayList
<
Integer
>
(list));

return
;
        } 
else
if
 (count 
>
 target) {

return
;
        }


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

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

continue
;
            }
            list.add(candidates[i]);
            search(candidates, target, i, list, result);
            list.remove(list.size() - 
1
);
        }
    }


private
int
sum
(List
<
Integer
>
 list)
{

if
 (list == 
null
) {

return
0
;
        }


int
 count = 
0
;

for
 (
int
 i = 
0
; i 
<
 list.size(); i++) {
            count += list.get(i);
        }


return
 count;
    }
}

九章算法提供的解答将list求和操作融合在递归函数中，感觉更简洁，供参考。

Java 实现

public
class
Solution
{

/**
     * 
@param
 candidates: A list of integers
     * 
@param
 target:An integer
     * 
@return
: A list of lists of integers
     */
public
 List
<
List
<
Integer
>
>
 combinationSum(
int
[] candidates, 
int
 target) {
        List
<
List
<
Integer
>
>
 result = 
new
 ArrayList
<
>
();

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

return
 result;
        }

        List
<
Integer
>
 list = 
new
 ArrayList
<
>
();
        Arrays.sort(candidates);
        search(candidates, target, 
0
, list, result);

return
 result;
    }


private
void
search
(
int
[] candidates,

int
 target,

int
 pos,
                        List
<
Integer
>
 list,
                        List
<
List
<
Integer
>
>
 result)
{

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

return
;
        }          


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

if
 (candidates[i] 
>
 target) {

break
;
            }


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

continue
;
            }

            list.add(candidates[i]);
            search(candidates, target - candidates[i], i, list, result);
            list.remove(list.size() - 
1
);
        }
    }
}

参考

1. Combination Sum | 九章算法