[array] leetcode - 40. Combination Sum II - Medium

leetcode - 40. Combination Sum II - Medium

descrition

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]
]

解析

和 leetcode - 39. Combination Sum - Medium 类似，只是这里的数组元素存在重复，并且元素不可重复取。

代码只实现了其中一种递归形式，这样的实现方法递归层数应该是最浅的。

code

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class Solution{
public:
    vector<vector<int> > combinationSum2(vector<int> &candidates, int target){
        vector<vector<int> > ans;
        vector<int> vecCur;
        sort(candidates.begin(), candidates.end());
        combinationSum2Backtracking(candidates, 0, vecCur, target, ans);
        return ans;
    }

    void combinationSum2Backtracking(vector<int>& candidates, int index,
                                     vector<int>& vecCur, int target,
                                     vector<vector<int> > &ans){
        if(target < 0)
            return;
        if(target == 0){
            if(!vecCur.empty())
                ans.push_back(vecCur);
            return;
        }

        for(int i=index; i<candidates.size(); i++){
            if(candidates[i] > target) // candidates mush in ascending order
                break;
            // choose candidates[i], and each number in candidates may only
            // be used onece in combination
            vecCur.push_back(candidates[i]);
            combinationSum2Backtracking(candidates, i+1, vecCur, target - candidates[i], ans);
            // don‘t choose candidates[i]
            vecCur.pop_back();

            // skip the duplicate
            while((i+1)<candidates.size() && candidates[i+1] == candidates[i])
                i++;
            // after i++, i will point to a new unique number
        }
    }
};

int main()
{
    return 0;
}