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【c++】【leetcode 40】组合总和II

曾宝月 2022-03-10 阅读 50

解题思路

回溯+去重
与组合问题差不多的做法,该问题需要去重
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代码

class Solution {
public:
    vector<vector<int>> ans;
    vector<int> path;

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        DFS(candidates, target, 0);
        return ans;
    }
    void DFS(vector<int> candidates, int target, int level) {
        if (target < 0)return;//这个可以不要,在for循环中有判断
        if (target == 0) {
            ans.push_back(path);
            return;
        }

        for (int i = level; i < candidates.size() && candidates[i] <= target; i++) {
            //如果进入了for循环,代表在同一层,同一层有用过直接跳过
            if(i > level && candidates[i] == candidates[i-1])continue;
            path.push_back(candidates[i]);
            DFS(candidates, target - candidates[i], i + 1);
            //进行回溯
            path.pop_back();
        }
        return;
    }
};
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