linear-list/array/3sum
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2023-12-01
3Sum
描述
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
- Elements in a triplet
(a,b,c)must be in non-descending order. (ie, $$a \leq b \leq c$$) - The solution set must not contain duplicate triplets.
For example, given array S = {-1 0 1 2 -1 -4}.
A solution set is:
(-1, 0, 1)
(-1, -1, 2)
分析
先排序，然后双指针左右夹逼，时间复杂度 $$O(n^2)$$。
这个方法可以推广到k-sum，先排序，然后做k-2次循环，在最内层循环左右夹逼，时间复杂度是 $$O(\max{n \log n, n^{k-1}})$$。
代码
双指针
# 3Sum
# 先排序，然后左右夹逼，注意跳过重复的数
# Time Complexity: O(n^2)
# Space Complexity: from O(logn) to O(n), depending on the
# implementation of the sorting algorithm
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
result = []
for i in range(len(nums)):
if i == 0 or nums[i] != nums[i-1]:
for lst in self.twoSumII(nums, i+1, 0-nums[i]):
result.append([nums[i]]+lst)
return result
def twoSumII(self, nums: List[int], start: int, target:int)->List[List[int]]:
result = []
low, high = start, len(nums)-1
while low < high:
sum = nums[low] + nums[high];
if sum < target:
low += 1
elif sum > target:
high -= 1
else:
result.append([nums[low], nums[high]])
low += 1
high -= 1
while low < high and nums[low] == nums[low-1]:
low += 1
while low < high and nums[high] == nums[high+1]:
high -= 1
return result
// 3Sum
// 先排序，然后左右夹逼，注意跳过重复的数
// Time Complexity: O(n^2)
// Space Complexity: from O(logn) to O(n), depending on the
// implementation of the sorting algorithm
public class Solution {
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length; ++i) {
if (i == 0 || nums[i] != nums[i-1]) {
for(List<Integer> lst: twoSumII(nums, i+1, 0-nums[i])) {
lst.add(0, nums[i]);
result.add(lst);
}
}
}
return result;
}
public List<List<Integer>> twoSumII(int[] nums, int start, int target) {
List<List<Integer>> result = new ArrayList<>();
int low = start, high = nums.length-1;
while (low < high) {
int sum = nums[low] + nums[high];
if (sum < target) {
++low;
} else if(sum > target) {
--high;
} else {
result.add(new ArrayList<>(Arrays.asList(nums[low++], nums[high--])));
while(low < high && nums[low] == nums[low-1]) ++low;
while(low < high && nums[high] == nums[high+1]) --high;
}
}
return result;
}
};
// 3Sum
// 先排序，然后双指针左右夹逼，注意跳过重复的数
// Time Complexity: O(n^2)
// Space Complexity: from O(logn) to O(n), depending on the
// implementation of the sorting algorithm
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
sort(begin(nums), end(nums));
vector<vector<int>> result;
for (int i = 0; i < nums.size(); ++i)
if (i == 0 || nums[i] != nums[i - 1]) {
for (auto& lst: twoSumII(nums, i+1, 0-nums[i])) {
lst.insert(lst.begin(), nums[i]);
result.push_back(lst);
}
}
return result;
}
vector<vector<int>> twoSumII(vector<int>& nums, int start, int target) {
vector<vector<int>> result;
int low = start, high = nums.size() - 1;
while (low < high) {
int sum = nums[low] + nums[high];
if (sum < target) {
++low;
} else if (sum > target) {
--high;
} else {
result.push_back({ nums[low++], nums[high--] });
while(low < high && nums[low] == nums[low-1]) ++low;
while(low < high && nums[high] == nums[high+1]) --high;
}
}
return result;
}
};
HashSet
其他代码完全一样，仅仅是 twoSumII()不同。
# 3Sum
# 先排序，然后twoSumII()用HashSet实现
# Time Complexity: O(n^2), Space Complexity: O(n)
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
result = []
for i in range(len(nums)):
if i == 0 or nums[i] != nums[i-1]:
for lst in self.twoSumII(nums, i+1, 0-nums[i]):
result.append([nums[i]]+lst)
return result
def twoSumII(self, nums: List[int], start: int, target:int)->List[List[int]]:
result = []
s = set()
for i in range(start, len(nums)):
if len(result) == 0 or result[-1][1] != nums[i]:
complement = target - nums[i]
if complement in s:
result.append([complement, nums[i]])
s.add(nums[i])
return result
// 3Sum
// 先排序，然后twoSumII()用HashSet实现
// Time Complexity: O(n^2), Space Complexity: O(n)
public class Solution {
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length; ++i) {
if (i == 0 || nums[i] != nums[i-1]) {
for(List<Integer> lst: twoSumII(nums, i+1, 0-nums[i])) {
lst.add(0, nums[i]);
result.add(lst);
}
}
}
return result;
}
public List<List<Integer>> twoSumII(int[] nums, int start, int target) {
List<List<Integer>> result = new ArrayList<>();
Set<Integer> s = new HashSet<>();
for (int i = start; i < nums.length; ++i) {
if (result.isEmpty() || result.get(result.size() - 1).get(1) != nums[i]) {
int complement = target - nums[i];
if (s.contains(complement)) {
result.add(new ArrayList<>(Arrays.asList(complement, nums[i])));
}
}
s.add(nums[i]);
}
return result;
}
};
// 3Sum
// 先排序，然后twoSumII()用HashSet实现
// Time Complexity: O(n^2), Space Complexity: O(n)
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
sort(begin(nums), end(nums));
vector<vector<int>> result;
for (int i = 0; i < nums.size(); ++i)
if (i == 0 || nums[i] != nums[i - 1]) {
for (auto& lst: twoSumII(nums, i+1, 0-nums[i])) {
lst.insert(lst.begin(), nums[i]);
result.push_back(lst);
}
}
return result;
}
vector<vector<int>> twoSumII(vector<int>& nums, int start, int target) {
vector<vector<int>> result;
unordered_set<int> s;
for (auto i = start; i < nums.size(); ++i) {
if (result.empty() || result.back()[1] != nums[i]) {
int complement = target - nums[i];
if (s.count(complement)) {
result.push_back({ complement, nums[i]});
}
}
s.insert(nums[i]);
}
return result;
}
};