[Backtracking] [leetcode] and the combination of the target number (the number can be reused)

topic:

Given an array candidates with no duplicate elements and a target number target, find out all combinations of candidates that can make the number sum target.

The numbers in candidates can be repeatedly selected without limitation.

Description:

• All numbers (including target) are positive integers.
• The solution set cannot contain repeated combinations. 

prompt:

• 1 <= candidates.length <= 30
• 1 <= candidates[i] <= 200
• Each element in the candidate is unique.
• 1 <= target <= 500

Example:

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

source:

39. Combination Sum

Problem-solving ideas: backtracking

Continue to learn the idea of ​​backtracking. The difference from the previous two combinations is that this question 1) the array is not sorted, so it is sorted first; 2) the number can be reused, so the parameter is no longer +1 during recursion.

• Recursive termination condition: current sum is greater than or equal to target
• The result meets the condition: sum == target
• Pruning condition: the current sum is about to exceed the target value

c++ code:

class Solution {
public:
    vector< vector<int> > result;
    vector<int> path;
    vector< vector<int> > combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end()); // 排序
        back(candidates, target, 0, 0);
        return result;
    }
    void back(const vector<int>& candidates, int target, int sum, int start) {
        if (sum == target) {
            result.push_back(path);
            return;
        }

        for (int i = start; i < candidates.size(); i++) {
            if (sum + candidates[i] > target) break; // 剪枝
            path.push_back(candidates[i]);
            sum += candidates[i];
            back(candidates, target, sum, i); // 因为可以重复使用数字，所以i而不是i+1
            sum -= candidates[i];
            path.pop_back();
        }
    }
};