【LeetCode】167. Two Sum II – Input Array Is Sorted 解答・解説【Python】

問題
1. 原文
2. 内容
解答
補足・参考・感想

問題

原文

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].

1
2
3

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].

1
2
3

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].

1
2
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] <= 1000

numbers is sorted in non-decreasing order.

-1000 <= target <= 1000

The tests are generated such that there is exactly one solution.

内容

※正しくない可能性があります。

解答

解答1：two pointer

class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        #左右のポインタを用意
        l,r = 0, len(numbers)-1
        #左のポインタが右のポインタの左側にある限り繰り返す
        while l<r:
            #左ポインタが指す要素と右ポインタが指す要素の合計をsに代入
            s = numbers[l] + numbers[r]
            #sがtaretと一致した場合
            if s== target:
                #返す
                return [l+1, r+1]
            #sがtarget未満の場合
            elif s < target:
                #左ポインタを進める
                l += 1
            #sがtarget以上の場合
            else:
                #右ポインタを左に進める
                r -= 1

class Solution:

def twoSum(self, numbers: List[int], target: int) -> List[int]:

#左右のポインタを用意

l,r = 0, len(numbers)-1

#左のポインタが右のポインタの左側にある限り繰り返す

while l<r:

#左ポインタが指す要素と右ポインタが指す要素の合計をsに代入

s = numbers[l] + numbers[r]

#sがtaretと一致した場合

if s== target:

#返す

return [l+1, r+1]

#sがtarget未満の場合

elif s < target:

#左ポインタを進める

l += 1

#sがtarget以上の場合

else:

#右ポインタを左に進める

r -= 1

解答2：ハッシュテーブル(辞書)

解答3：二分探索

補足・参考・感想

■補足

■参考

■感想

前：283. Move Zeroes

次：344. Reverse String

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