Combination Sum II —— LeetCode

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

10,1,2,7,6,1,5

and target

8

,
A solution set is:

[1, 7]

[1, 2, 5]

[2, 6]

[1, 1, 6]

题目大意：跟上一题类似，但是有点区别就是候选集中的元素只能出现一次。

解题思路：每次从当前元素的下一个开始计算sum，并把candidate加入List，并且跳过重复元素。

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<>();
if (candidates == null || candidates.length == 0) {
return res;
}
Deque<Integer> tmp = new ArrayDeque<>();
Arrays.sort(candidates);
helper(res, tmp, 0, target, candidates);
return res;
}

private void helper(List<List<Integer>> res, Deque<Integer> tmp, int start, int target, int[] candidate) {
if (target == 0) {
res.add(new ArrayList<>(tmp));
//            System.out.println(tmp);
return;
}
for (int i = start; i < candidate.length && target >= candidate[i]; i++) {
tmp.addLast(candidate[i]);
helper(res, tmp, i + 1, target - candidate[i], candidate);
tmp.removeLast();
while (i < candidate.length - 1 && candidate[i + 1] == candidate[i]) {
i++;
}
}
}