leetcode: Combination Sum II

http://oj.leetcode.com/problems/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.

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：

如果Combination Sum解决了，那么这题只要改一行代码，因为每个数字只能出现一次。

class Solution {
public:
    void internalCombinationSum2(vector<int> &num,
            int start,
            int sum,
            int target,
            vector<int> &combination,
            vector<vector<int> > &result) {
        int size = num.size();
        
        if (sum == target) {
            result.push_back(combination);
            
            return;
        }
        else if ((start >= size) || (sum > target)) {
            return;
        }

        for (int i = start; i < size; ) {
            combination.push_back(num[i]);
            internalCombinationSum2(num, i + 1, sum + num[i], target, combination, result);
            combination.pop_back();
            
            int j = i + 1;
            
            while (j < size) {
                if (num[i] == num[j]) {
                    ++j;
                }
                else {
                    break;
                }
            }
            
            i = j;
        }
    }
    
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        vector<vector<int> > result;
        vector<int> combination;
        
        sort(num.begin(), num.end());
        internalCombinationSum2(num, 0, 0, target, combination, result);
           
        return result;
    }
};