[LeetCode] 40. Combination Sum II 组合之和 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]
]

39. Combination Sum的变形，39题数组中的数字可以重复使用，而这道题数组中的数字不能重复使用。这里要考虑跳过重复的数字，其它的与39题一样。

解法：和39一样，递归 + backtracking

Java:

public List<List<Integer>> combinationSum2(int[] cand, int target) {
    Arrays.sort(cand);
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    List<Integer> path = new ArrayList<Integer>();
    dfs_com(cand, 0, target, path, res);
    return res;
}
void dfs_com(int[] cand, int cur, int target, List<Integer> path, List<List<Integer>> res) {
    if (target == 0) {
        res.add(new ArrayList(path));
        return ;
    }
    if (target < 0) return;
    for (int i = cur; i < cand.length; i++){
        if (i > cur && cand[i] == cand[i-1]) continue;
        path.add(path.size(), cand[i]);
        dfs_com(cand, i+1, target - cand[i], path, res);
        path.remove(path.size()-1);
    }
}　　

Java:

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
    List<List<Integer>> result = new ArrayList<List<Integer>>();
    List<Integer> curr = new ArrayList<Integer>();
    Arrays.sort(candidates);
    helper(result, curr, 0, target, candidates);
    return result;
}

public void helper(List<List<Integer>> result, List<Integer> curr, int start, int target, int[] candidates){
    if(target==0){
        result.add(new ArrayList<Integer>(curr));
        return;
    }
    if(target<0){
        return;
    }

    int prev=-1;
    for(int i=start; i<candidates.length; i++){
        if(prev!=candidates[i]){ // each time start from different element
            curr.add(candidates[i]);
            helper(result, curr, i+1, target-candidates[i], candidates); // and use next element only
            curr.remove(curr.size()-1);
            prev=candidates[i];
        }
    }
}　　

Python:

class Solution:
    # @param candidates, a list of integers
    # @param target, integer
    # @return a list of lists of integers
    def combinationSum2(self, candidates, target):
        result = []
        self.combinationSumRecu(sorted(candidates), result, 0, [], target)
        return result

    def combinationSumRecu(self, candidates, result, start, intermediate, target):
        if target == 0:
            result.append(list(intermediate))
        prev = 0
        while start < len(candidates) and candidates[start] <= target:
            if prev != candidates[start]:
                intermediate.append(candidates[start])
                self.combinationSumRecu(candidates, result, start + 1, intermediate, target - candidates[start])
                intermediate.pop()
                prev = candidates[start]
            start += 1

C++:

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        vector<vector<int> > res;
        vector<int> out;
        sort(num.begin(), num.end());
        combinationSum2DFS(num, target, 0, out, res);
        return res;
    }
    void combinationSum2DFS(vector<int> &num, int target, int start, vector<int> &out, vector<vector<int> > &res) {
        if (target < 0) return;
        else if (target == 0) res.push_back(out);
        else {
            for (int i = start; i < num.size(); ++i) {
                if (i > start && num[i] == num[i - 1]) continue;
                out.push_back(num[i]);
                combinationSum2DFS(num, target - num[i], i + 1, out, res);
                out.pop_back();
            }
        }
    }
};　　

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[LeetCode] 39. Combination Sum 组合之和