Question
Given an array of integers, 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.
Input: numbers={2, 7, 11, 15}, target=9 Output: index1=1, index2=2
Solution
- Brute Force (binary search)
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
struct Item {
int index;
int value;
bool operator<(const Item& a) const {
return value < a.value;
}
};
vector<int> twoSum(vector<int>& nums, int target) {
// put nums into an temp array with index stored
vector<Item> temp(nums.size());
for (int i = 0; i < nums.size(); i++) {
Item item;
item.index = i;
item.value = nums[i];
temp[i] = item;
}
// sort the array
sort(temp.begin(), temp.end());
// find result indexes
vector<int> result(2);
for (int i = 0; i < temp.size(); i++) {
Item dest;
dest.value = target - temp[i].value;
// find dest number by binary search
pair<vector<Item>::iterator, vector<Item>::iterator> range;
range = equal_range(temp.begin() + i + 1, temp.end(), dest);
vector<Item>::iterator lower = range.first;
vector<Item>::iterator upper = range.second;
if (lower != upper) {
result[0] = temp[i].index + 1; // non zero-based
result[1] = lower->index + 1;
if (result[0] > result[1]) {
swap(result[0], result[1]); // make result ordered
}
break;
}
}
return result;
}
};
- Two Pointers Search
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
struct Item {
int index;
int value;
bool operator<(const Item& a) const {
return value < a.value;
}
};
vector<int> twoSum(vector<int>& nums, int target) {
// put nums into an temp array with index stored
vector<Item> temp(nums.size());
for (int i = 0; i < nums.size(); i++) {
Item item;
item.index = i;
item.value = nums[i];
temp[i] = item;
}
// sort the array
sort(temp.begin(), temp.end());
// two pointer search
int left = 0;
int right = temp.size() - 1;
int sum = temp[left].value + temp[right].value;
while ((sum != target) && (left < right)) {
if (sum < target) {
left++;
} else {
right--;
}
sum = temp[left].value + temp[right].value;
}
// result founded when sum == target
vector<int> result(2);
result[0] = temp[left].index + 1; // non zero-based
result[1] = temp[right].index + 1;
if (result[0] > result[1]) { // make result ordered
swap(result[0], result[1]);
}
return result;
}
};