[LeetCode] 40. Combination Sum II #40

grandyang · 2019-05-30T16:14:09Z

请点击下方图片观看讲解视频
 Click below image to watch YouTube Video

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target.

Each number in candidates may only be used once in the combination.

Note: The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8
Output: 
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5
Output: 
[
[1,2,2],
[5]
]

Constraints:

1 <= candidates.length <= 100
1 <= candidates[i] <= 50
1 <= target <= 30

这道题跟之前那道 Combination Sum 本质没有区别，只需要改动一点点即可，之前那道题给定数组中的数字可以重复使用，而这道题不能重复使用，只需要在之前的基础上修改几个地方即可，首先要给数组排个序，然后在递归的 for 循环里加上 if (i > start && num[i] == num[i - 1]) continue; 这样可以防止 res 中出现重复项，最后就在递归调用 dfs 里面的参数换成 i+1，这样就不会重复使用数组中的数字了，代码如下：

解法一：

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        vector<int> cur;
        sort(candidates.begin(), candidates.end());
        dfs(candidates, target, 0, cur, res);
        return res;
    }
    void dfs(vector<int>& candidates, int target, int start, vector<int>& cur, vector<vector<int>>& res) {
        if (target < 0) return;
        if (target == 0) { res.push_back(cur); return; }
        for (int i = start; i < candidates.size(); ++i) {
            if (i > start && candidates[i] == candidates[i - 1]) continue;
            cur.push_back(candidates[i]);
            dfs(candidates, target - candidates[i], i + 1, cur, res);
            cur.pop_back();
        }
    }
};

对于之前的解法二可以通过稍微改动而适用于这里，同样的在处理当前数字 candidates[i] 时，和之前的数字比较，要跳过重复数字。其次就是在为下次递归创建新数组时，不能包括当前的数字，这样的话才能保证不重复使用数字，参见代码如下：

解法二：

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        sort(candidates.begin(), candidates.end());
        for (int i = 0; i < candidates.size(); ++i) {
            if (candidates[i] > target) break;
            if (candidates[i] == target) { res.push_back({candidates[i]}); break; }
            if (i > 0 && candidates[i] == candidates[i - 1]) continue;
            vector<int> vec = vector<int>(candidates.begin() + i + 1, candidates.end());
            vector<vector<int>> tmp = combinationSum2(vec, target - candidates[i]);
            for (auto a : tmp) {
                a.insert(a.begin(), candidates[i]);
                res.push_back(a);
            }
        }
        return res;
    }
};

对于 DP 解法来说，改变就比较大了，甚至可以说是完全不同的解法也不为过，这种原数组中有重复数字，且每个数字只能使用一次的要求，不太适合用 DP 来做，但也可以强行使用 DP 来做，只不过稍微有点麻烦。

解法三：

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<vector<int>>> dp(target + 1);
        dp[0].resize(1);
        map<int, int> numCnt;
        for (int num : candidates) {
            ++numCnt[num];
        }
        for (auto a : numCnt) {
            int num = a.first, cnt = a.second;
            for (int i = target - num; i >= 0; --i) {
                for (auto v : dp[i]) {
                    int sum = i;
                    for (int k = 0; k < cnt && sum <= target - num; ++k) {
                        sum += num;
                        v.push_back(num);
                        dp[sum].push_back(v);
                    }
                }
            }
        }
        return dp[target];
    }
};

Github 同步地址：

#40

类似题目：

Combination Sum III

Combination Sum

参考资料：

https://leetcode.com/problems/combination-sum-ii/

https://leetcode.com/problems/combination-sum-ii/discuss/16861/Java-solution-using-dfs-easy-understand

https://leetcode.com/problems/combination-sum-ii/discuss/16878/Combination-Sum-I-II-and-III-Java-solution-(see-the-similarities-yourself)

LeetCode All in One 题目讲解汇总(持续更新中...)

（欢迎加入博主的知识星球，博主将及时答疑解惑，并分享刷题经验与总结，快快加入吧～）

喜欢请点赞，疼爱请打赏❤️~.~