组合总和

[题目](39. 组合总和)

代码实现

class Solution {
private:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(vector<int> & canditates, int target, int sum, int startIndex) {
        if (sum == target) {
            result.push_back(path);
            return;
        }
        // 如果sum + candidates[i] > target 就终止遍历
        for(int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++) {
            sum += candidates[i];
            path.push_back(candidates[i]);
            // 不用i + 1，表示可以重复读取当前的数字
            backtracking(candidates, target, sum, i);
            sum -= candidates[i];
            path.pop_back();
        }
    }
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        backtracking(candidates, target, 0, 0);
        return result;
    }
};

组合总和II

[题目](40. 组合总和 II)

重点

需要事先进行排序，从小到大

代码实现

class Solution {
private:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(vector<int> & canditates, int target, int sum, int startIndex, vector<bool> & used) {
        if (sum == target) {
            result.push_back(path);
            return;
        }
        for(int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++) {
            // used[i - 1] == true，说明同一树枝candidates[i - 1]使用过
            // used[i - 1] == false，说明同一树层candidates[i - 1]使用过
            // 要对同一树层使用过的元素进行跳过
            if (i > 0 && candidates[i] == candidates[i - 1] && used[i - 1] == false) {
                continue;
            }
            sum += candidates[i];
            path.push_back(candidates[i]);
            used[i] = true;
            backtracking(candidates, target, sum, i + 1, used);
            used[i] = false;
            sum -= candidates[i];
            path.pop_back();
        }
    }
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<bool> used(candidates.size(), false);
        // 首先排序
        sort(candidates.begin(), candidates.end());
        backtracking(candidates, target, 0, 0, used);
        return result;
    }
};