[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(qū)別，只需要改動一點點即可，之前那道題給定數(shù)組中的數(shù)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(huán)里加上 if (i > start && num[i] == num[i - 1]) continue; 這樣可以防止 res 中出現(xiàn)重復項，然后就在遞歸調用 helper 里面的參數(shù)換成 i+1，這樣就不會重復使用數(shù)組中的數(shù)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();
        }
    }
};

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