LeetCode-040 Combination Sum II

Description

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates 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.

Example

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]

Analyse

这是上一道题的变式，将可以重复选的数字改成了所有的数字只能使用一次，方法同样是DFS，但在这里，必须得进行排序（前一道题排不排序都可以），因为数组里会出现两个或多个相同数字的情况，取的时候，必须把他们的区别性消除掉，否则就会出现这种情况，拿Example 2举个例子，为了区分，我们把相同的2设为2_a,2_b,2_c，那么[1,2,2]就会出现[1,2_a,2_b]，[1,2_a,2_c]，[1,2_b,2_c]，三种情况，这样是不对的，我们可以通过排序来解决这个问题。出现多个的情况，我们只采用最后一个作为代表。

Code

class Solution {
public:
    void calcombination(vector<vector<int>> &ans, vector<int> &temp, int target, vector<int>& candidates, int begin) {
        if (target == 0) {
            ans.push_back(temp);
        } else {
            for (int i = begin; i < candidates.size(); i++) {
                if (candidates[i] > target) break;
                if (i != begin && candidates[i] == candidates[i - 1]) continue;
                temp.push_back(candidates[i]);
                calcombination(ans, temp, target - candidates[i], candidates, i + 1);
                temp.pop_back();
            }
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> ans;
        vector<int> temp;
        calcombination(ans, temp, target, candidates, 0);
        return ans;
    }
};