從數(shù)組中尋找和的相加數(shù)

Astrian 發(fā)布于2019-07-25 12:05 / 1026人閱讀

摘要：從數(shù)組中尋找和的相加數(shù)思路這個題目簡單，兩個循環(huán)嵌套可以實現(xiàn)，但是考慮到輸入數(shù)組長度大的情況下，時間復雜度太高使用索引，遍歷一次后記錄中所有數(shù)字出現(xiàn)的下標，用得到，然后直接查索引中的位置索引可以用數(shù)組，但是中可能會有大數(shù)

從數(shù)組中尋找和的相加數(shù) Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

example 1

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

example 2

Given nums = [0,4,3,0], target = 0,

Because nums[0] + nums[3] = 0 + 0 = 0,
return [0, 3].

example 3

Given nums = [-1,-2,-3,-4,-5], target = -8,

Because nums[2] + nums[4] = -3 + (-5) = -8,
return [2, 4].

example 4

Given nums = [0,1,2,3,4,5,...,28888,...,65432], target = 65432,

you should consider that time and space is limited.

思路

這個題目簡單，兩個for循環(huán)嵌套可以實現(xiàn)，但是考慮到輸入數(shù)組長度大的情況下，時間復雜度太高, O(n2)

使用索引，遍歷一次后記錄nums中所有數(shù)字出現(xiàn)的下標，用target - nums[i]得到j(luò)，然后直接查索引中j的位置

索引可以用list(數(shù)組)，但是nums中可能會有大數(shù)字，所需的空間太多，并且會有負數(shù)，改用dir，key為nums[i]，value為i

同時給出簡單版代碼（簡單版在java，C等語言在leetcode網(wǎng)站可以AC，但python會超時）

代碼

優(yōu)化版

class Solution:

    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        idx_set = {}
        for i, item in enumerate(nums):
            idx = target - nums[i]
            if idx in idx_set:
                j = idx_set[idx]
                if len(j) != 0 and j[0] != i:
                    return list([i, j[0]])
                if j[0] == i and len(j) > 1:
                    return list([i, j[1]])
            else:
                idx_set[item] = [i]

簡單版

class Solution:

    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for i in range(len(nums)):
            for j in range(i+1, len(nums)):
                if nums[i] + nums[j] == target:
                    return list([i, j])

本題以及其它leetcode題目代碼github地址: github地址