- Leetcode 1. Two Sum
- Leetcode 167. Two Sum II - Input array is sorted
- Leetcode 15. 3Sum
- LeetCode 16. 3Sum Closest
- LeetCode 18. 4Sum
Leetcode Sum Serials
Leetcode 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.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
The idea: use the map to store the value’s index, and for loop the map to find m[target-nums[i]] exits in the map
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> res;
const int n = nums.size();
unordered_map<int, int> m;
for (int i=0; i<n; i++)
m[nums[i]] = i;
for (int i=0; i<n; i++) {
if (m[target-nums[i]]) {
res.push_back(i);
res.push_back(m[target-nums[i]]);
break;
}
}
return res;
}
};
Leetcode 167. Two Sum II - Input array is sorted
Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution and you may not use the same element twice.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int left=0, right = numbers.size()-1;
while (left<right) {
if (numbers[left]+numbers[right]==target)
return {left + 1, right + 1};
else if (numbers[left]+numbers[right]<target) left++;
else right--;
}
}
};
Leetcode 15. 3Sum
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: 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]
]
The idea: sort the nums, and then start at i, j=i+1, and k is the last number, if the sum is smaller, then j++, if it is bigger, then k–;
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> res;
const int n = nums.size();
if (n<3) return res;
sort(nums.begin(), nums.end());
for (int i=0; i<n; i++) {
int j = i+1, k = n-1;
while (j<k) {
while (j<k && nums[i]+nums[j]+nums[k] < 0) j++;
while (j<k && nums[i]+nums[j]+nums[k] > 0) k--;
if (j<k && nums[i]+nums[j]+nums[k] == 0) {
res.push_back({nums[i], nums[j], nums[k]});
while (j<k && nums[j]==nums[j+1]) j++;
j++;
while (j<k && nums[k]==nums[k-1]) k--;
}
}
while (i<n && nums[i]==nums[i+1]) i++;
}
return res;
}
};
LeetCode 16. 3Sum Closest
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
The idea: sort the nums, keep trace the newest closet answer and difference.
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int ans = nums[0]+nums[1]+nums[2];
int diff = abs(target - ans);
if (diff==0) return ans;
const int n = nums.size();
sort(nums.begin(), nums.end());
for (int i=0; i<n; i++) {
int j = i+1;
int k = n-1;
while (j<k) {
int temp = nums[i]+nums[j]+nums[k];
int newDiff = abs(target-temp);
if (newDiff<diff) {
ans = temp;
diff = newDiff;
}
if (temp<target) j++;
else k--;
}
}
return ans;
}
};
LeetCode 18. 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: 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]
]
The idea is same as the 3Sum.
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> res;
const int n = nums.size();
if (n<4) return res;
sort(nums.begin(), nums.end());
for (int i=0; i<n; i++) {
for (int j=i+1; j<n; j++) {
int k = j+1;
int m = n-1;
while (k<m) {
while (k<m && nums[i]+nums[j]+nums[k]+nums[m] < target) k++;
while (k<m && nums[i]+nums[j]+nums[k]+nums[m] > target) m--;
if (k<m && nums[i]+nums[j]+nums[k]+nums[m] == target) {
res.push_back({nums[i], nums[j], nums[k], nums[m]});
while (k<m && nums[k]==nums[k+1]) k++;
k++;
while (k<m && nums[m]==nums[m-1]) m--;
}
}
while (j<n && nums[j]==nums[j+1]) j++;
}
while (i<n && nums[i]==nums[i+1]) i++;
}
return res;
}
};