40. Combination Sum II

最新推荐文章于 2025-05-25 18:32:47 发布

原创最新推荐文章于 2025-05-25 18:32:47 发布 · 224 阅读

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本文解析了一道经典的算法题目“组合求和II”的解决方案，通过递归的方式寻找所有可能的组合，并通过排序和去重操作确保结果的唯一性。

1刷
跟上一题基本一样，就是递归的时候变成不重复了，最后要去重！！！
PS
sort(last.begin(), last.end());
vector

class Solution {
public:
    void getit(int target, int sum, int num, vector<int>& cand, vector<int>& now, vector<vector<int>>& ll){
     if(sum == target){
         ll.push_back(now);
         return;
     }   
     if(sum > target) return;

     for(int i = num; i < cand.size(); ++ i){
         sum += cand[i];
         now.push_back(cand[i]);
         getit(target, sum, i + 1, cand, now, ll);
         now.pop_back();
         sum -= cand[i];
     }
     return;
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> last;
        vector<int> now;
         getit(target, 0, 0, candidates, now, last);
         sort(last.begin(), last.end());
         vector<vector<int>>::iterator it;
         it = unique(last.begin(), last.end());
         if(it != last.end()) last.erase(it, last.end());
         return last;
    }
};

2刷
跟上一题基本一样，就是加了一句去重复的，还有i+1
if(i > 0 && cand[i - 1] == cand[i] && i > begin) continue;

class Solution {
public:
    vector<vector<int>>ve;
    void findit(vector<int>& cand, vector<int> & vec, int target, int begin){
        if(target == 0){
            ve.push_back(vec);
            return ;
        }
        for(int i = begin; i < cand.size() && target >= cand[i]; ++ i){
            if(i > 0 && cand[i - 1] == cand[i] && i > begin) continue;
            vec.push_back(cand[i]);
            findit(cand, vec, target - cand[i], i + 1);
            vec.pop_back();
        }

    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<int>vec;
        findit(candidates, vec, target, 0);
        return ve;
    }
};