[LeetCode] 40. Combination Sum II 組合之和之二

 

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]
]

 

這道題跟之前那道 Combination Sum 本質沒有區別，只需要改動一點點即可，之前那道題給定數組中的數字可以重復使用，而這道題不能重復使用，只需要在之前的基礎上修改兩個地方即可，首先在遞歸的 for 循環里加上 if (i > start && num[i] == num[i - 1]) continue; 這樣可以防止 res 中出現重復項，然后就在遞歸調用 helper 里面的參數換成 i+1，這樣就不會重復使用數組中的數字了，代碼如下：

 

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

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/40

 

類似題目：

Combination Sum III

Combination Sum

 

參考資料：

https://leetcode.com/problems/combination-sum-ii/

https://leetcode.com/problems/combination-sum-ii/discuss/16861/Java-solution-using-dfs-easy-understand

https://leetcode.com/problems/combination-sum-ii/discuss/16878/Combination-Sum-I-II-and-III-Java-solution-(see-the-similarities-yourself)

 

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