leetcode40:Combination Sum II
阿新 • • 發佈:2018-12-11
思路:本題在於數字不能重複使用,而且需要去重,可以先將陣列排序,然後去重就方便了,依然採用深度優先的方法。
程式碼:
public class CombinationSumII { public static void main(String[] args) { int[] num = { 2, 5, 2, 1, 2 }; System.out.println(combinationSum(num, 5)); } public static List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> ls = new ArrayList<>(); List<Integer> list = new ArrayList<>(); Arrays.sort(candidates); helpCombinationSum(candidates, target, 0, ls, list); return ls; } private static void helpCombinationSum(int[] num, int target, int index, List<List<Integer>> ls, List<Integer> list) { if (target == 0 && !ls.contains(list))// 此處注意contains和containsAll的區別 ls.add(new ArrayList(list)); if (index >= num.length || target < 0)//注意此處等號,如果放在上面,需要去掉等號 return; for (int i = index; i < num.length; i++) { list.add(num[i]); helpCombinationSum(num, target - num[i], i + 1, ls, list); list.remove(list.size() - 1); } } }
輸出:
附:
contains:判斷list1中是否包含list2物件;
containsAll:判斷list1中是否包含list2中所有元素。