Combination Sum
描述
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 () must be in non-descending order. (ie, ).
- 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]
分析
无
代码
// Combination Sum
// 时间复杂度O(n!),空间复杂度O(n)
public class Solution {
public List<List<Integer>> combinationSum(int[] nums, int target) {
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>(); // 最终结果
List<Integer> path = new ArrayList<>(); // 中间结果
dfs(nums, path, result, target, 0);
return result;
}
private static void dfs(int[] nums, List<Integer> path,
List<List<Integer>> result, int gap, int start) {
if (gap == 0) { // 找到一个合法解
result.add(new ArrayList<Integer>(path));
return;
}
for (int i = start; i < nums.length; i++) { // 扩展状态
if (gap < nums[i]) return; // 剪枝
path.add(nums[i]); // 执行扩展动作
dfs(nums, path, result, gap - nums[i], i);
path.remove(path.size() - 1); // 撤销动作
}
}
}