2014
11-19

# LeetCode-4Sum[数组]

### 4Sum

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, abcd)
• The solution set must not contain duplicate quadruplets.
    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

A solution set is:
(-1,  0, 0, 1)
(-2, -1, 1, 2)
(-2,  0, 0, 2)


// LeetCode, 4Sum
// 先排序，然后左右夹逼，时间复杂度O(n^3)，空间复杂度O(1)
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& num, int target) {
vector<vector<int>> result;
if (num.size() < 4) return result;
sort(num.begin(), num.end());

auto last = num.end();
for (auto a = num.begin(); a < prev(last, 3); ++a) {
for (auto b = next(a); b < prev(last, 2); ++b) {
auto c = next(b);
auto d = prev(last);
while (c < d) {
if (*a + *b + *c + *d < target) {
++c;
} else if (*a + *b + *c + *d > target) {
--d;
} else {
result.push_back({ *a, *b, *c, *d });
++c;
--d;
}
}
}
}
sort(result.begin(), result.end());
result.erase(unique(result.begin(), result.end()), result.end());
return result;
}
};


// LeetCode, 4Sum
// 用一个hashmap先缓存两个数的和
// 时间复杂度，平均O(n^2)，最坏O(n^4)，空间复杂度O(n^2)
class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target) {
vector<vector<int>> result;
if (num.size() < 4) return result;
sort(num.begin(), num.end());

unordered_map<int, vector<pair<int, int> > > cache;
for (size_t a = 0; a < num.size(); ++a) {
for (size_t b = a + 1; b < num.size(); ++b) {
cache[num[a] + num[b]].push_back(pair<int, int>(a, b));
}
}

for (int c = 0; c < num.size(); ++c) {
for (size_t d = c + 1; d < num.size(); ++d) {
const int key = target - num[c] - num[d];
if (cache.find(key) == cache.end()) continue;

const auto& vec = cache[key];
for (size_t k = 0; k < vec.size(); ++k) {
if (c <= vec[k].second)
continue; // 有重叠

result.push_back( { num[vec[k].first],
num[vec[k].second], num[c], num[d] });
}
}
}
sort(result.begin(), result.end());
result.erase(unique(result.begin(), result.end()), result.end());
return result;
}
};


// LeetCode, 4Sum
// 用一个 hashmap 先缓存两个数的和
// 时间复杂度O(n^2)，空间复杂度O(n^2)
// @author 龚陆安(http://weibo.com/luangong)
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& num, int target) {
vector<vector<int>> result;
if (num.size() < 4) return result;
sort(num.begin(), num.end());

unordered_multimap<int, pair<int, int>> cache;
for (int i = 0; i + 1 < num.size(); ++i)
for (int j = i + 1; j < num.size(); ++j)
cache.insert(make_pair(num[i] + num[j], make_pair(i, j)));

for (auto i = cache.begin(); i != cache.end(); ++i) {
int x = target - i->first;
auto range = cache.equal_range(x);
for (auto j = range.first; j != range.second; ++j) {
auto a = i->second.first;
auto b = i->second.second;
auto c = j->second.first;
auto d = j->second.second;
if (a != c && a != d && b != c && b != d) {
vector<int> vec = { num[a], num[b], num[c], num[d] };
sort(vec.begin(), vec.end());
result.push_back(vec);
}
}
}
sort(result.begin(), result.end());
result.erase(unique(result.begin(), result.end()), result.end());
return result;
}
};


// LeetCode, 4Sum
// 先排序，然后左右夹逼，时间复杂度O(n^3logn)，空间复杂度O(1)，会超时
// 跟方法1相比，表面上优化了，实际上更慢了，切记！
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& num, int target) {
vector<vector<int>> result;
if (num.size() < 4) return result;
sort(num.begin(), num.end());

auto last = num.end();
for (auto a = num.begin(); a < prev(last, 3);
a = upper_bound(a, prev(last, 3), *a)) {
for (auto b = next(a); b < prev(last, 2);
b = upper_bound(b, prev(last, 2), *b)) {
auto c = next(b);
auto d = prev(last);
while (c < d) {
if (*a + *b + *c + *d < target) {
c = upper_bound(c, d, *c);
} else if (*a + *b + *c + *d > target) {
d = prev(lower_bound(c, d, *d));
} else {
result.push_back({ *a, *b, *c, *d });
c = upper_bound(c, d, *c);
d = prev(lower_bound(c, d, *d));
}
}
}
}
return result;
}
};


1. 在方法1里面：

//遍历所有的边，计算入度
for(int i=0; i<V; i++)
{
degree = 0;