leecode-39. Combination Sum
1、問題描述:
Given a set of candidate numbers (C) (without duplicates) 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] ]
2、邊界條件:無
3、思路:先取一個數,然後與target比較;==則存貯,!=則繼續從所有數裏面選擇。
從Level-N裏面選擇一個數,與目標比較,符合則存貯;不符合則再從所有數裏面挨個取,從而化為同樣的Level-N問題
形成遞歸。base case:1)與目標匹配;2)target - nums[i]<0,再減下去也是負數,這依賴於題目給的 全正數positive integers 條件
4、實現代碼:
class Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> results = new ArrayList<>(); //Arrays.sort(candidates); combinationSum(results, new ArrayList<Integer>(), candidates, target);return results; } public void combinationSum(List<List<Integer>> results, List<Integer> cur, int[] candidates, int target) { if (0 == target) {
/** Collections.sort(cur);//這裏排序會把原列表改變,所以上層在恢復現場時出錯。 if (!results.contains(cur)) {//去重 results.add(new ArrayList<Integer>(cur)); }
**/
ArrayList<Integer> result = new ArrayList<Integer>(cur);//先生成新的cur,然後進行排序
Collections.sort(result); //
if (!results.contains(result)) {
results.add(result);
return;
} if (0 > target) { return; } for (int i = 0; i < candidates.length; i++) { cur.add(candidates[i]); combinationSum(results, cur, candidates, target - candidates[i]); cur.remove(cur.size() - 1); } } }
5、時間復雜度:說不好; 空間復雜度:
6、題外知識:Arraylist排序:Collections靜態排序API,Collections的排序都是穩定的。Collections.sort(List<T> list)、和Collections.sort(List<T> list,Comparator<?super T> c);使用的排序是穩定的,主要是對list排序。
鏈接:http://blog.csdn.net/tuke_tuke/article/details/51100219 和 http://www.importnew.com/17211.html
leecode-39. Combination Sum