40. Combination Sum II

Posted 2020-09-29

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.
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]
]

比39. Combination Sum多了个从下一位递归而已

public List<List<Integer>> combinationSum2(int[] num, int target) {
        // write your code here
       public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        //List<Integer> list = new ArrayList<Integer>();
        
        if (candidates.length == 0) {
            return res;
        }
        Arrays.sort(candidates);
        helper(res, new ArrayList<Integer>(), candidates, target, 0);
        return res;
    }
    
    private void helper(List<List<Integer>> res, List<Integer> list,
                        int[] num, int target, int pos) {
        if (target == 0) {
            res.add(new ArrayList<Integer>(list));
            return;
        }
        for (int i = pos; i < num.length; i++) {
            if (num[i] > target || i != pos && num[i] == num[i - 1]) {
                continue;
            }
            list.add(num[i]);
            helper(res, list, num, target - num[i], i + 1);
            list.remove(list.size() - 1);
            
        }
    }

　利用排序的性质 提高效率　

if (num[i] > target || i != pos && num[i] == num[i - 1]) {
                continue;
            }