4Sum
描述
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, ) - 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)
分析
先排序,然后左右夹逼,复杂度 ,会超时。
可以用一个hashmap先缓存两个数的和,最终复杂度。这个策略也适用于 3Sum 。
左右夹逼
// 4Sum
// 先排序,然后左右夹逼
// Time Complexity: O(n^3),Space Complexity: O(1)
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> result;
if (nums.size() < 4) return result;
sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size() - 3; ++i) {
if (i > 0 && nums[i] == nums[i-1]) continue;
for (int j = i + 1; j < nums.size() - 2; ++j) {
if (j > i+1 && nums[j] == nums[j-1]) continue;
int k = j + 1;
int l = nums.size() - 1;
while (k < l) {
const int sum = nums[i] + nums[j] + nums[k] + nums[l];
if (sum < target) {
++k;
while(nums[k] == nums[k-1] && k < l) ++k;
} else if (sum > target) {
--l;
while(nums[l] == nums[l+1] && k < l) --l;
} else {
result.push_back({nums[i], nums[j], nums[k], nums[l]});
++k;
--l;
while(nums[k] == nums[k-1] && k < l) ++k;
while(nums[l] == nums[l+1] && k < l) --l;
}
}
}
}
return result;
}
};
HashMap 做缓存
// 4Sum
// 用一个hashmap先缓存两个数的和
// Time Complexity: 平均O(n^2),最坏O(n^4),Space Complexity: O(n^2)
class Solution {
public:
vector<vector<int> > fourSum(vector<int> &nums, int target) {
vector<vector<int>> result;
if (nums.size() < 4) return result;
sort(nums.begin(), nums.end());
unordered_map<int, vector<pair<int, int> > > cache;
for (size_t a = 0; a < nums.size(); ++a) {
for (size_t b = a + 1; b < nums.size(); ++b) {
cache[nums[a] + nums[b]].push_back(pair<int, int>(a, b));
}
}
for (int c = 0; c < nums.size(); ++c) {
for (size_t d = c + 1; d < nums.size(); ++d) {
const int key = target - nums[c] - nums[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( { nums[vec[k].first],
nums[vec[k].second], nums[c], nums[d] });
}
}
}
sort(result.begin(), result.end());
result.erase(unique(result.begin(), result.end()), result.end());
return result;
}
};