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算法渣，现实基本都参考或者完全拷贝[戴方勤(soulmachine@gmail.com)]大神的leetcode题解，此处仅作刷题记录。
1 class Solution { 2 public: 3 vector<vector<int> > fourSum(vector<int> &num, int target) { 4 vector<vector<int> > result; 5 if (num.size() < 4) 6 return result; 7 sort(num.begin(), num.end()); 8 9 unordered_map<int, vector<pair<int, int> > > cache; 10 for (size_t a = 0; a < num.size(); a++){ 11 for (size_t b = a + 1; b < num.size(); b++) { 12 cache[num[a] + num[b]].push_back(pair<int, int>(a, b)); // 对于某个和，可能存在多种组合，这些组合存储到一个向量中 13 } 14 } 15 16 for (size_t c = 0; c < num.size(); c++) { 17 for (size_t d = c + 1; d < num.size(); d++) { 18 const int key = target - num[c] - num[d]; 19 if (cache.find(key) == cache.end()) 20 continue; 21 22 const auto& vec = cache[key]; 23 for (size_t k = 0; k < vec.size(); ++k) { 24 if (c <= vec[k].second) // 重复 25 continue; 26 result.push_back({ num[vec[k].first], num[vec[k].second], num[c], num[d] }); 27 } 28 } 29 } 30 31 sort(result.begin(), result.end()); 32 result.erase(unique(result.begin(), result.end()), result.end()); 33 return result; 34 } 35 };
下面方法感觉更好
1 class Solution { 2 public: 3 vector<vector<int>> fourSum(vector<int>& num, int target) { 4 vector<vector<int>> result; 5 if (num.size() < 4) return result; 6 sort(num.begin(), num.end()); 7 unordered_multimap<int, pair<int, int>> cache; 8 for (int i = 0; i + 1 < num.size(); ++i) 9 for (int j = i + 1; j < num.size(); ++j) 10 cache.insert(make_pair(num[i] + num[j], make_pair(i, j))); 11 for (auto i = cache.begin(); i != cache.end(); ++i) { 12 int x = target - i->first; 13 auto range = cache.equal_range(x); 14 for (auto j = range.first; j != range.second; ++j) { 15 auto a = i->second.first; 16 auto b = i->second.second; 17 auto c = j->second.first; 18 auto d = j->second.second; 19 if (a != c && a != d && b != c && b != d) { 20 vector<int> vec = { num[a], num[b], num[c], num[d] }; 21 sort(vec.begin(), vec.end()); 22 result.push_back(vec); 23 } 24 } 25 } 26 sort(result.begin(), result.end()); 27 result.erase(unique(result.begin(), result.end()), result.end()); 28 return result; 29 } 30 };
杂记:
1. pair
This class couples together a pair of values, which may be of different types (T1 and T2). The individual values can be accessed through its public members first and second.
Pairs are a particular case of tuple.
2. 其他方法用到的类 unordered_multimap
Unordered multimaps are associative containers that store elements formed by the combination of a key value and amapped value, much like unordered_map containers, but allowing different elements to have equivalent keys.
Elements with equivalent keys are grouped together in the same bucket and in such a way that an iterator (seeequal_range) can iterate through all of them.
3. insert
unordered_multimap没有[]运算符，必须以insert方式插入数据，而且插入式必须调用make_pair函数
4. equal_range
Returns the bounds of the subrange that includes all the elements of the range [first,last) with values equivalent tova
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原文地址：http://www.cnblogs.com/Azurewing/p/4308660.html
