1. Two Sum
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
法一: 利用排序+对撞指针，时间复杂度O(nlogn),空间复杂度O(1)
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> res;
int i = 0, j = nums.size() - 1;
while(left < right)
{
if(nums[i] + nums[j] == target)
{
res.push_back({i, j});
break;
}
else if(nums[i] + nums[j] < target)
i ++;
else
j --;
}
return res;
}法二: 空间换时间，时间复杂度O(n),空间复杂度O(n)
vector<int> twoSum(vector<int>& nums, int target) {
map<int, int> mp;
vector<int> res;
for(int i = 0; i < nums.size(); i++)
{
if(mp.find(target - nums[i]) != mp.end())
{
res= {mp[target- nums[i]], i};
break;
}
else
mp[nums[i]] = i;
}
return res;
}15. 3Sum
Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
The solution set must not contain duplicate triplets.
法：外层循环+2sum，注意去重，时间复杂度O(n^2),空间复杂度O(1)
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> res;
if(nums.size() < 3)
return res;
sort(nums.begin(), nums.end());
for(int i = 0; i < nums.size() - 2; i ++)
{
if(i > 0 && nums[i] == nums[i - 1])
continue;
int left = i + 1, right = nums.size() - 1;
while(left < right)
{
if(nums[left] + nums[right] == -nums[i])
{
vector<int> result = {nums[i], nums[left ++], nums[right --]};
res.push_back(result);
while(left < right && nums[left] == nums[left-1])
left++;
while(left < right && nums[right] == nums[right+1])
right--;
}
else if(nums[left] + nums[right] > -nums[i])
right --;
else
left ++;
}
}
return res;
}16. 3Sum Closest
Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
法：与3sum类似，可以不去重，时间复杂度O(n^2),空间复杂度O(1)
int threeSumClosest(vector<int>& nums, int target) {
int res = 0;
int mgap = INT_MAX;
sort(nums.begin(), nums.end());
for(int i = 0; i < nums.size() - 2; i ++)
{
int j = i + 1;
int k = nums.size() - 1;
while(j < k)
{
int gap = abs(nums[i] + nums[j] + nums[k] - target);
if(gap < mgap)
{
res = nums[i] + nums[j] + nums[k];
mgap = gap;
}
if(nums[i] + nums[j] + nums[k] < target)
j ++;
else
k --;
}
}
return res;
}18. 4Sum
Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a+ b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
The solution set must not contain duplicate quadruplets.
法：双层层循环+2sum，注意去重，时间复杂度O(n^3),空间复杂度O(1)
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> res;
if(nums.size() < 3)
return res;
sort(nums.begin(), nums.end());
for(int j = 0; j <nums.size() - 3; j ++)
{
if(j > 0 && nums[j] == nums[j - 1])
continue;
for(int i = j + 1; i < nums.size() - 2; i ++)
{
if(i > j + 1 && nums[i] == nums[i - 1])
continue;
int left = i + 1, right = nums.size() - 1;
while(left < right)
{
if(nums[left] + nums[right] == target-nums[i]-nums[j])
{
vector<int> result = {nums[j], nums[i], nums[left ++], nums[right --]};
res.push_back(result);
while(left < right && nums[left] == nums[left-1])
left++;
while(left < right && nums[right] == nums[right+1])
right--;
}
else if(nums[left] + nums[right] > target-nums[i]-nums[j])
right --;
else
left ++;
}
}
}
return res;
}454. 4Sum II
Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.
To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 2^31 - 1.
法：2sum+2sum，记录2sum的值，时间复杂度O(n^2),空间复杂度O(n^2)
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
if(A.size() == 0)
return 0;
int res = 0;
int N = A.size();
unordered_map<int, int> mp;
for(int i = 0; i < N; i ++)
for(int j = 0; j < N; j++)
mp[A[i] + B[j]] ++;
for(int i = 0; i < N; i ++)
for(int j = 0; j < N; j ++)
{
if(mp.find(-C[i]-D[j]) != mp.end())
res += mp[-C[i]-D[j]];
}
return res;
}总结
1.排序后前后指针对撞。
2.空间换时间，用hashmap存储中间结果、