40. Combination Sum II

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates 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.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

Solution

(1) Java

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> rst = new ArrayList<>();
        if (candidates == null || candidates.length == 0) {
            return rst;
        }
        Arrays.sort(candidates);
        Set<List<Integer>>[] dp = new Set[target+1];
        for (int i = 0; i <= target; i++) {
            dp[i] = new HashSet<List<Integer>>();
        }
        for (int j = 0; j < candidates.length; j++) {
            int c = candidates[j];
            for (int i = target; i >= 1; i--) {
                if (c == i) {
                    List<Integer> l = new ArrayList<Integer>();
                    l.add(c);
                    dp[i].add(l);                    
                } else if (c < i) {
                    Set<List<Integer>> l = dp[i-c];
                    for (List<Integer> ll : l) {
                        List<Integer> lll = new ArrayList<>(ll);
                        lll.add(c);
                        dp[i].add(lll);
                    }
                }
            }
        }
        for(List<Integer> l : dp[target]) {
            rst.add(l);
        }
        return rst;
    }
}

(2) Python

class Solution:
    def combinationSum2(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        rst = []
        if not candidates:
            return rst
        candidates.sort()
        self.helper(candidates, target, rst, [], 0)
        return rst

    def helper(self, candidates, target, rst, path, start):
        if target == 0:
            path2 = list(path)
            rst.append(path2)
            return
        elif target < 0:
            return
        for i in range(start, len(candidates)):
            if i != start and candidates[i] == candidates[i-1]:
                continue
            path.append(candidates[i])
            self.helper(candidates, target-candidates[i], rst, path, i+1)
            path.pop()

(3) Scala