#1. Two Sum
Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order.
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
if nums[j] == target - nums[i]:
return [i, j]
def twoSum(self, nums, target):
seen = {}
for i, v in enumerate(nums):
remaining = target - v
if remaining in seen:
return [seen[remaining], i]
seen[v] = i
return []#88. Merge Sorted Array
You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.
Merge nums1 and nums2 into a single array sorted in non-decreasing order.
The final sorted array should not be returned by the function, but instead, be stored inside the array nums1. To accommodate this, nums1 has a length of m+n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.
class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""
# Set p1 and p2 to point to the end of their respective arrays.
p1 = m - 1
p2 = n - 1
# And move p backwards through the array, each time writing
# the smallest value pointed at by p1 or p2.
for p in range(n + m - 1, -1, -1):
if p2 < 0:
break
if p1 >= 0 and nums1[p1] > nums2[p2]:
nums1[p] = nums1[p1]
p1 -= 1
else:
nums1[p] = nums2[p2]
p2 -= 1