LeetCode Combination Sum II

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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.
• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• 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] 

思路：稍微改一下就行了，一个是每次的下标都+1，另外就是避免重复，跳过相同的数

class Solution {
public:
    void dfs(vector<int> candidates, int index, int sum, int target, vector<vector<int>> &res, vector<int> &path) {
        if (sum > target) return;
        if (sum == target) {
            res.push_back(path);
            return;
        }
        for (int i = index; i < candidates.size(); i++) {
            path.push_back(candidates[i]);
            dfs(candidates, i+1, sum+candidates[i], target, res, path);
            path.pop_back();
            while (i < candidates.size() -1 && candidates[i] == candidates[i+1]) i++;
        }
    }
    vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int> > res;
        vector<int> path;
        dfs(candidates, 0, 0, target, res, path);
        return res;
    }
};

LeetCode Combination Sum II

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原文地址：http://blog.csdn.net/u011345136/article/details/44057273