学步园

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]

关键是判断重复，如果两个数相同，只有第一个数使用了，才能使用第二个数。

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        sort(num.begin(), num.end());
        
        vector<int> v;
        combinationSumUtil(num, v, target, 0);
        
        return result;
    }
    
    void combinationSumUtil(vector<int> &candidates, vector<int> &comb, int sum, int start) {
        if (sum == 0) {
            result.push_back(comb);
            return;
        }
        
        for (int i = start; i < candidates.size(); ++i) {
            if (candidates[i] <= sum) {
                if (!isDuplicate(candidates, comb, i)) {
                    comb.push_back(candidates[i]);
                    combinationSumUtil(candidates, comb, sum - candidates[i], i + 1);
                    comb.pop_back();
                }
            }
            else {
                break;
            }
        }
    }
    
    bool isDuplicate(vector<int> &candidates, vector<int> &comb, int index) {
        int same = 0;
        int i = index;
        while (--i >= 0) {
            if (candidates[i] == candidates[i+1]) {
                ++same;
            }
            else {
                break;
            }
        }
        i = comb.size();
        while (--i >= 0) {
            if (comb[i] == candidates[index]) {
                --same;
            }
            else {
                break;
            }
        }
        
        return (same > 0);
    }

private:
    vector<vector<int> > result;
};

也可以通过每次出栈的时候跳过相同元素来避免重复。

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        sort(num.begin(), num.end());
        
        vector<int> v;
        combinationSumUtil(num, v, target, 0);
        
        return result;
    }
    
    void combinationSumUtil(vector<int> &candidates, vector<int> &comb, int sum, int start) {
        if (sum == 0) {
            result.push_back(comb);
            return;
        }
        
        for (int i = start; i < candidates.size(); ++i) {
            if (candidates[i] <= sum) {
                comb.push_back(candidates[i]);
                combinationSumUtil(candidates, comb, sum - candidates[i], i + 1);
                comb.pop_back();
                while (i + 1 < candidates.size() && candidates[i] == candidates[i+1]) {
                    ++i;
                }
            }
            else {
                break;
            }
        }
    }

private:
    vector<vector<int> > result;
};