Combination Sum II

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.

http://www.lintcode.com/en/problem/combination-sum-ii/

Discussion

Combination Sum 问题的扩展，跟Subsets II问题一样的方法去充就可以了。
if(i!=start && nums[i] == nums[i-1]) continue; -- 这种情况下就跳过当前number。

Solution

class Solution {
public:
    /**
     * @param num: Given the candidate numbers
     * @param target: Given the target number
     * @return: All the combinations that sum to target
     */
    vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
        // write your code here
        vector<vector<int> > result;
        int n = candidates.size();
        if(n==0) return result;
        vector<int> solution;
        sort(candidates.begin(), candidates.end());//忘记一百次！！！
        dfs(candidates, target, solution, 0, result);
        return result;
    }
    void dfs(vector<int> &candidates, int gap, vector<int> &solution, int start, 
    vector<vector<int> > &result) {
        if(gap == 0) {
            result.push_back(solution);
            return;
        }
        int n = candidates.size();
        for(int i=start; i<n; i++) {
            if(gap < candidates[i]) {
                break;
            }
            if(i != start && candidates[i] == candidates[i-1])
                continue;
            solution.push_back(candidates[i]);
            dfs(candidates, gap-candidates[i], solution, i+1, result);
            solution.pop_back();
        }
    }
};
`