LeetCode_1.Two Sum

问题

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:

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

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

解题

一般思路：

class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        # Time Limit Exceeded
        for i in range(len(nums) - 1):
            for j in range(i + 1, len(nums)):
                if nums[i] + nums[j] == target:
                    return [i, j]

以上方法在LeetCode中运行会超时:-(

改进方式：

class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        dict = {}
        for i in range(len(nums)):
            if target - nums[i] in dict:
                return [dict[target - nums[i]], i]
            else:
                dict[nums[i]] = i

时间复杂度从O(n² )降至O(n)。

思路

利用字典数据类型的特性，减少额外的循环操作。