leetcode 40 combination Sum II

最新推荐文章于 2025-08-17 23:28:34 发布

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最新推荐文章于 2025-08-17 23:28:34 发布

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本文介绍了一种在给定候选数集合与目标数的情况下找出所有可能的唯一组合的方法。每个候选数只能使用一次，且解决方案中不能包含重复的组合。通过具体的示例展示了如何找到所有可能的组合。

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

Each number in C 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.
For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]

class Solution {
public:
    typedef vector<int>::iterator It;

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        It begin = candidates.begin();
        It end = candidates.end();

        sort(begin, end);
        combinate(begin, end, target);

        return ret;
    }

private:
    void combinate(It begin, It end, int target) {

        for (It it = begin; it != end; ++it) {
            int val = *it;
            if (val < target) {
                ans.push_back(val);
                combinate(it + 1, end, target -val);
                ans.pop_back();
            } else if (val == target) {
                ans.push_back(val);
                ret.push_back(ans);
                ans.pop_back();
                return;
            } else {
                return;
            }

            // 防止重复寻找
            while(it + 1 != end && *it == *(it + 1)){
                ++it;
            }
        }
    }
    vector<vector<int> > ret;
    vector<int> ans;
};