【算法题】组合总和 II

给定一个候选人编号的集合 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的每个数字在每个组合中只能使用 一次 。

注意：解集不能包含重复的组合。 

示例 1:

输入: candidates = [10,1,2,7,6,1,5], target = 8,
输出:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
示例 2:

输入: candidates = [2,5,2,1,2], target = 5,
输出:
[
[1,2,2],
[5]
]
 

提示:

1 <= candidates.length <= 100
1 <= candidates[i] <= 50
1 <= target <= 30

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        List<Integer> trace = new ArrayList<>();
        backtrace(candidates, target, 0, trace, 0);
        return res;
    }

    public void backtrace(int[] candidates, int target, int start, List<Integer> trace, int sum){
        if(sum == target){
            res.add(new ArrayList<Integer>(trace));
            return;
        }

        for(int i = start; i < candidates.length; i++){
            // 避免重复
            if(i > start && candidates[i] == candidates[i-1]){
                continue;
            }
            if(candidates[i] + sum <= target){
                trace.add(candidates[i]);
                backtrace(candidates, target, i + 1, trace, sum + candidates[i]);
                trace.remove(trace.size() - 1);
            }else{
                // 后面的 candidates[i] 都比当前大
                break;
            }
        }
    }
}