[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();

            }

        }

    }

};

类似题目：

[LeetCode] 39. Combination Sum 组合之和

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

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