Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
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
10,1,2,7,6,1,5and target8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
和上一题思路一样，只是条件不同了。private void getCom(int[] candidates, int target, int start, List<Integer> subList, List<List<Integer>> list) { for (int i = start; i < candidates.length; i++) { if (i == start || candidates[i] != candidates[i - 1]) { List<Integer> newList = new ArrayList<Integer>(subList); if (target - candidates[i] == 0) { newList.add(candidates[i]); list.add(newList); } else if (target - candidates[i] > 0) { newList.add(candidates[i]); getCom(candidates, target - candidates[i], i + 1, newList, list); } else { break; } } } } public List<List<Integer>> combinationSum2(int[] num, int target) { Arrays.sort(num); List<List<Integer>> list = new ArrayList<List<Integer>>(); List<Integer> subList = new ArrayList<Integer>(); getCom(num, target, 0, subList, list); return list; }
Combination Sum II,布布扣,bubuko.com
原文地址：http://blog.csdn.net/u010378705/article/details/30304961
