Given an array of integers numbers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.
Return the indices of the two numbers (1-indexed) as an integer array answer of size 2, where 1 <= answer[0] < answer[1] <= numbers.length.
You may assume that each input would have exactly one solution and you may not use the same element twice.
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.
Example 2:
Input: numbers = [2,3,4], target = 6 Output: [1,3]
Example 3:
Input: numbers = [-1,0], target = -1 Output: [1,2]
Constraints:
2 <= numbers.length <= 3 * 104-1000 <= numbers[i] <= 1000numbersis sorted in increasing order.-1000 <= target <= 1000- Only one valid answer exists.
Solution 1: (Two pointer)
/**
* @param {number[]} numbers
* @param {number} target
* @return {number[]}
*/
var twoSum = function(numbers, target) {
let l = 0;
let r = numbers.length - 1;
while (l < r) {
let sum = numbers[l] + numbers[r];
if (sum === target) break;
else if (sum > target) r--;
else l++;
}
// this question use 1-indexed
return [l + 1, r + 1];
};Solution 2: (Hash table)
var twoSum = function(numbers, target) {
let map = new Map;
for (var i = 0; i < numbers.length; i++) {
var diff = target - numbers[i];
if(map.has(diff)) {
return [map.get(diff), i + 1];
}
map.set(numbers[i], i + 1);
}
};