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.
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]
]
我想好半天也没把递归搞清楚,在网上看了一个比较好的算法,跟着学了一遍,debug跟着递归走一走,感觉拨开迷雾一样。
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if (candidates.length == 0)
return res;
List<Integer> temp = new ArrayList<Integer>();
Arrays.sort(candidates);
findSum(candidates, target, 0, 0, temp, res);
return res;
}
private void findSum(int[] candidates, int target, int sum, int level,
List<Integer> temp, List<List<Integer>> res) {
if (sum == target) {
res.add(new ArrayList<Integer>(temp));
return;
} else if (sum > target)
return;
else
for (int i = level; i < candidates.length; i++) {
temp.add(candidates[i]);
findSum(candidates, target, sum + candidates[i], i, temp, res);
temp.remove(temp.size() - 1);
}
}