Here‘s no Duke anymore

minty_Brit666

Today’s blog is about the practice of the leetcode.
And I’ll give my own answer in this blog.

1. The sum of two numbers
   **Given an integer array nums and an integer target value target, find the two integers in the array and the target value target and return their array subscripts.
   You can assume that there is only one answer for each type of input. However, the same element in the array cannot be repeated in the answer.
   You can return the answers in any order.

Source: LeetCode
Link: https://leetcode-cn.com/problems/two-sum
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**
we can use the for loop twice to iterate through all the values and find the two numbers that sum to the target values:

def twoSum(self, nums, target):
    n=len(nums)
    for idx in range(n):
        for jdx in range(idx+1,n):
            if nums[idx] + nums[jdx] == target:
                return idx,jdx

Or we can use the index to reduce a loop:

def twoSum(self, nums, target):
    for idx in range(len(nums)):
        if target - nums[idx] in nums:
            if idx != nums.index(target - nums[idx]):
                return idx,nums.index(target - nums[idx])

The easiest way to do that is to use a hash dictionary:

def twoSum(self, nums, target):
    dict = {}
    for idx in range(len(nums)):
        x = target - nums[idx]
        if x in dict:
            return dict[x],idx
        else:
            dict[nums[idx]] = idx

Of the three methods above, the last one consumes the least amount of system resources and takes the least amount of time.