40. 组合总和 II(leetcode)

https://leetcode.cn/problems/combination-sum-ii/description/

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    LinkedList<Integer> path = new LinkedList<>();
    int sum = 0;
    
    public List<List<Integer>> combinationSum2(int[] candidates,int target) {
        //为了将重复的数字都放到一起，所以先进行排序
        Arrays.sort( candidates );
        backTracking( candidates, target, 0 );
        return res;
    }
    
    private void dfs(int[] candidates,int target,int start) {
        if(sum == target) {
            res.add( new ArrayList<>( path ) );
            return;
        }
        for( int i = start; i < candidates.length && sum + candidates[i] <= target; i++ ) {
            //正确剔除重复解的办法
            //跳过同一树层使用过的元素,因为前一次搜索已经搜索完以这个元素开头的组合了
            if ( i > start && candidates[i] == candidates[i - 1] ) {
                continue;
            }
            sum += candidates[i];
            path.add(candidates[i]);
            // i+1 代表当前组内元素只选取一次
            dfs(candidates,target,i + 1);
            int temp = path.getLast();
            sum -= temp;
            path.removeLast();
        }
    }
}