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.
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 2,3,6,7 and target 7,
A solution set is: [7] [2, 2, 3]
解题思路：
首先说明下题目Bug，实际测试中C中元素是不会重复的，因此降低了不少难度，最简单的实现方法，DFS算法，JAVA实现如下：
static public List<List<Integer>> combinationSum(int[] candidates,int target) {
List<List<Integer>> list = new ArrayList<List<Integer>>();
Arrays.sort(candidates);
dfs(list, candidates, 0, target, 0);
return list;
}
static List<Integer> list2 = new ArrayList<Integer>();
static void dfs(List<List<Integer>> list, int[] array, int result,int target, int depth) {
if (result == target) {
list.add(new ArrayList<Integer>(list2));
return;
}
else if (depth >= array.length || result > target)
return;
for (int i = 0; i <= target / array[depth]; i++) {
for (int j = 0; j < i; j++)
list2.add(array[depth]);
dfs(list, array, result + array[depth] * i, target, depth+1);
for (int j = 0; j < i; j++)
list2.remove(list2.size() - 1);
}
}