40. Combination Sum II

 

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 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]
]

 

这道题跟之前那道 Combination Sum 本质没有区别,只需要改动一点点即可,之前那道题给定数组中的数字可以重复使用,而这道题不能重复使用,只需要在之前的基础上修改两个地方即可,首先在递归的 for 循环里加上 if (i > start && num[i] == num[i - 1]) continue; 这样可以防止 res 中出现重复项,然后就在递归调用 helper 里面的参数换成 i+1,这样就不会重复使用数组中的数字了,代码如下:

 

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& num, int target) {
        vector<vector<int>> res;
        vector<int> out;
        sort(num.begin(), num.end());
        helper(num, target, 0, out, res);
        return res;
    }
    void helper(vector<int>& num, int target, int start, vector<int>& out, vector<vector<int>>& res) {
        if (target < 0) return;
        if (target == 0) { res.push_back(out); return; }
        for (int i = start; i < num.size(); ++i) {
            if (i > start && num[i] == num[i - 1]) continue;
            out.push_back(num[i]);
            helper(num, target - num[i], i + 1, out, res);
            out.pop_back();
        }
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/40

 

类似题目:

Combination Sum III

Combination Sum

 

参考资料:

https://leetcode.com/problems/combination-sum-ii/

https://leetcode.com/problems/combination-sum-ii/discuss/16861/Java-solution-using-dfs-easy-understand

https://leetcode.com/problems/combination-sum-ii/discuss/16878/Combination-Sum-I-II-and-III-Java-solution-(see-the-similarities-yourself)

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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