Javascript(leetcode#167) Two Sum II - Input Array Is Sorted

Difficult:Medium

題目

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.

翻譯

給定一個 1 索引的整數數組，該數組已經按非遞減順序排序，找到兩個數字，使它們加起來等於特定的目標數字。讓這兩個數字為 numbers[index1] 和 numbers[index2]，其中 1 <= index1 < index2 <= numbers.length。

返回兩個數字 index1 和 index2 的索引，加一作為長度為 2 的整數數組 [index1, index2]。

生成的測試使得只有一種解決方案。您不能兩次使用相同的元素。

您的解決方案只能使用恆定的額外空間。

範例

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].

解題思路

1. 利用obj將sub及num的位置存入
2. 比較差值若找到回傳當前位置及差值位置

程式碼

var twoSum = function(numbers, target) {
    let obj={};

    for(let i=0;i<numbers.length;i++){
      let sub=target-numbers[i];
      if(!obj[sub]){
        obj[numbers[i]]=i+1;
      }
      else{
        return  [obj[sub],i+1];
      }
    }
    return obj;
};