Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.
Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2.
The tests are generated such that there is exactly one solution. You may not use the same element twice.
Your solution must use only constant extra space.
Example 1:
Input: numbers = [2,7,11,15], target = 9 Output: [1,2] Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].
Example 2:
Input: numbers = [2,3,4], target = 6 Output: [1,3] Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].
Example 3:
Input: numbers = [-1,0], target = -1 Output: [1,2] Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].
Constraints:
2 <= numbers.length <= 3 * 104-1000 <= numbers[i] <= 1000numbersis sorted in non-decreasing order.-1000 <= target <= 1000- The tests are generated such that there is exactly one solution.
Binary search:
class Solution:
def twoSum(self, numbers: List[int], target: int) -> List[int]:
n = len(numbers)
for i in range(n - 1):
x = target - numbers[i]
j = bisect.bisect_left(numbers, x, lo=i + 1)
if j != n and numbers[j] == x:
return [i + 1, j + 1]
return [-1, -1]Two pointers:
class Solution:
def twoSum(self, numbers: List[int], target: int) -> List[int]:
i, j = 1, len(numbers)
while i < j:
x = numbers[i - 1] + numbers[j - 1]
if x == target:
return [i, j]
if x < target:
i += 1
else:
j -= 1
return [-1, -1]Binary search:
class Solution {
public int[] twoSum(int[] numbers, int target) {
for (int i = 0, n = numbers.length; i < n - 1; ++i) {
int x = target - numbers[i];
int left = i + 1, right = n - 1;
while (left < right) {
int mid = (left + right) >> 1;
if (numbers[mid] >= x) {
right = mid;
} else {
left = mid + 1;
}
}
if (numbers[left] == x) {
return new int[]{i + 1, left + 1};
}
}
return new int[]{-1, -1};
}
}Two pointers:
class Solution {
public int[] twoSum(int[] numbers, int target) {
int i = 1, j = numbers.length;
while (i < j) {
int x = numbers[i - 1] + numbers[j - 1];
if (x == target) {
return new int[] {i, j};
}
if (x < target) {
++i;
} else {
--j;
}
}
return new int[] {-1, -1};
}
}Binary search:
function twoSum(numbers: number[], target: number): number[] {
for (let i = 0, n = numbers.length; i < n - 1; ++i) {
const x = target - numbers[i];
let left = i + 1,
right = n - 1;
while (left < right) {
const mid = (left + right) >> 1;
if (numbers[mid] >= x) {
right = mid;
} else {
left = mid + 1;
}
}
if (numbers[left] == x) {
return [i + 1, left + 1];
}
}
return [-1, -1];
}Two pointers:
function twoSum(numbers: number[], target: number): number[] {
let i = 1,
j = numbers.length;
while (i < j) {
const x = numbers[i - 1] + numbers[j - 1];
if (x == target) {
return [i, j];
}
if (x < target) {
++i;
} else {
--j;
}
}
return [-1, -1];
}Binary search:
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
for (int i = 0, n = numbers.size(); i < n - 1; ++i) {
int x = target - numbers[i];
int j = lower_bound(numbers.begin() + i + 1, numbers.end(), x) - numbers.begin();
if (j != n && numbers[j] == x) return {i + 1, j + 1};
}
return {-1, -1};
}
};Two pointers:
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int i = 1, j = numbers.size();
while (i < j)
{
int x = numbers[i - 1] + numbers[j - 1];
if (x == target) return {i, j};
if (x < target) ++i;
else --j;
}
return {-1, -1};
}
};Binary search:
func twoSum(numbers []int, target int) []int {
for i, n := 0, len(numbers); i < n-1; i++ {
x := target - numbers[i]
left, right := i+1, n-1
for left < right {
mid := (left + right) >> 1
if numbers[mid] >= x {
right = mid
} else {
left = mid + 1
}
}
if numbers[left] == x {
return []int{i + 1, left + 1}
}
}
return []int{-1, -1}
}Two pointers:
func twoSum(numbers []int, target int) []int {
i, j := 1, len(numbers)
for i < j {
x := numbers[i-1] + numbers[j-1]
if x == target {
return []int{i, j}
}
if x < target {
i++
} else {
j--
}
}
return []int{-1, -1}
}Binary search:
/**
* @param {number[]} numbers
* @param {number} target
* @return {number[]}
*/
var twoSum = function (numbers, target) {
for (let i = 0, n = numbers.length; i < n - 1; ++i) {
const x = target - numbers[i];
let left = i + 1,
right = n - 1;
while (left < right) {
const mid = (left + right) >> 1;
if (numbers[mid] >= x) {
right = mid;
} else {
left = mid + 1;
}
}
if (numbers[left] == x) {
return [i + 1, left + 1];
}
}
return [-1, -1];
};Two pointers:
/**
* @param {number[]} numbers
* @param {number} target
* @return {number[]}
*/
var twoSum = function (numbers, target) {
let i = 1,
j = numbers.length;
while (i < j) {
const x = numbers[i - 1] + numbers[j - 1];
if (x == target) {
return [i, j];
}
if (x < target) {
++i;
} else {
--j;
}
}
return [-1, -1];
};use std::cmp::Ordering;
impl Solution {
pub fn two_sum(numbers: Vec<i32>, target: i32) -> Vec<i32> {
let n = numbers.len();
let mut l = 0;
let mut r = n - 1;
loop {
match (numbers[l] + numbers[r]).cmp(&target) {
Ordering::Less => l += 1,
Ordering::Greater => r -= 1,
Ordering::Equal => break,
}
}
vec![l as i32 + 1, r as i32 + 1]
}
}