算法分析与设计——LeetCode:39. Combination Sum

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本文介绍了一种解决组合求和问题的算法实现。通过递归地遍历候选数集合，寻找所有可能的组合使得这些数的总和等于目标值。算法采用深度优先搜索策略，并通过剪枝避免重复组合。

题目

Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:

[
  [7],
  [2, 2, 3]
]

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
    }
};

思路

这是从题目的讨论中看到的方法，循环遍历，如[2, 3, 6, 7]，当i为0，放入一个2进入combination，再继续在[2, 3, 6, 7]里选择；当i为1时，则不再有2，只在[3, 6, 7]里选择，最终可以涵盖所有情况。

代码

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int> > res;
        vector<int> combination;
        combinationSum(candidates, target, res, combination, 0);
        return res;
    }
    void combinationSum(vector<int> &candidates, int target, vector<vector<int> > &res, vector<int> &combination, int head) {
        if (target == 0) {
            res.push_back(combination);
            return;
        }
        for (int i = head; i < candidates.size() && candidates[i] <= target; i++) {
            combination.push_back(candidates[i]);
            combinationSum(candidates, target - candidates[i], res, combination, i);
            combination.pop_back();
        }
    }
};