[問題] leecode167 Time Limit Exceeded

[Kuba4ma ()]

題目
Given an array of integers that is already sorted in ascending order, 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.

Note:

Your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution and you may
not use the same element twice.
Example:

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.




code:
class Solution(object):
    def twoSum(self, numbers, target):
        """
        :type numbers: List[int]
        :type target: int
        :rtype: List[int]
        """
        answer=list()
        length=len(numbers)
        for i in range(length):
            for j in range(i+1,length):
                if numbers[i]+numbers[j]==target:
                    answer.append(i+1)
                    answer.append(j+1)
                    return answer





問題:
runcode沒問題，但submit會Time Limit Exceeded，是leecode沒辦法接受時間複雜度太高
的答案嗎?


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推 yushes920179: 純回答你問題 是
推 cuteSquirrel: 對，Time Complexity太高。
推 cuteSquirrel: 提示:O(n)演算法、利用以排序的性質、double index
→ cuteSquirrel:                       已排序
推 jn8029: 討論區有O(n)的解法啊
推 cuteSquirrel: 對，討論區upvote按讚數高的解法，很值得一讀。