39. Combination Sum

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每日一题 2019 - 03 - 26

题目：

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

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

Note:

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

Example 1:

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Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

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Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

解法：

这个题让我们在给定的数组里面找到与 target 值相等的数字和的序列，简单来说就是递归回溯的思想，从数组里面一个数字出发，深层次遍历所有的分支，也即：

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if(sum == target) {
    res.push_back(ans); 
    return; 
}
else if(sum > target)
    return;
else{
    for(int i = start; i < nums_len; i++){
        ans.push_back(candidates[i]);
        dfs(i, sum + candidates[i], target, candidates);
        ans.pop_back();
    }
}

这个题的解题关键是把回溯的条件要设置清楚，以及要把迭代过程中的暂存 vector 弹出

代码：

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class Solution {
public:
    vector<vector<int>>res;
    vector<int>ans;
    int nums_len;

    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        if(candidates.empty()) return res;
        nums_len = candidates.size();
        dfs(0, 0, target, candidates);
        return res;
    }

    void dfs(int start, int sum, int target,vector<int> candidates){
        if(sum == target) {
            res.push_back(ans); return; }
        else if(sum > target)
            return;
        else{
            for(int i = start; i < nums_len; i++){
                ans.push_back(candidates[i]);
                dfs(i, sum + candidates[i], target, candidates);
                ans.pop_back();
            }
        }
    }
};
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