LeetCode: Combination Sum II

最新推荐文章于 2019-03-06 12:45:47 发布

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最新推荐文章于 2019-03-06 12:45:47 发布

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分类专栏： LeetCode 文章标签： LeetCode array combination sum II c语言

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本文介绍了一种寻找候选数集合中所有唯一组合的算法，这些组合的和等于目标数。该算法避免了重复组合，并确保所有数值均为正整数。

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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 {
    private:
    int index_count;
    vector<vector<int> > results;
    
public:
    void backtrace(int target, int sum, vector<int>& candidates, vector<int>& index, int id, int n)
    {
        if (sum > target) {
            return;
        }
        if (sum == target)
        {
            vector<int> result;
            for (int i = 1; i <= n; ++i)
            {
                result.push_back(candidates[index[i]]);
            }
            results.push_back(result);
            return;
        }
        
        // To avoid repeat 
        for (int i = id + 1; i < candidates.size(); ++i)
        {
            index[n+1] = i;
            backtrace(target, sum + candidates[i], candidates, index, i, n+1);
            while(i < candidates.size() && candidates[i] == candidates[i+1]) {   // remove repeat numbers
                i++;
            }
            
        }
    }
    
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        
        vector<int> real_cdts;
        for (int i = 0; i < candidates.size(); ++i) {
            if (candidates[i] <= target) {
                real_cdts.push_back(candidates[i]);
            }
            else {
                break;
            }
        }
        
        index_count = 10000;
        vector<int> index = vector<int>(index_count, 0);
        
        results.clear();
        backtrace(target, 0, real_cdts, index, -1, 0);
        
        return results;
    }
};