LeetCode:1. Two Sum

原文链接

Question

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].

My Solution

Train of thought:

1. Actually, we only need to process a half of the given nums, the ones larger than half of the target or the ones smaller than that. As the above example, if we process nums smaller than half of the target, we can just judge whether 7 is in the list or not when processing the number 2 and ignore those larger than half of the target.

What you should pay more attention to: there’s just one solution, and you can use one specific element only once (cases like ([3, 2, 4], 6) and ([3, 3, 4], 6), you may get the wrong answer[0, 0]).

class MySolution(object):
def twoSum(self, nums, target):
half = target / 2
index1 = index2 = -1

for i, n in enumerate(nums):
if n <= half:
try:
index2 = nums.index(target - n)
if index2 != i:
index1 = i
else:
index1 = index2
index2 = nums.index(half, index1 + 1)
break

except ValueError:
pass

return [index1, index2]

Editorial Solution of LeetCode

Train of thought:

Reduce the look up time by trading space for speed. The code is more succinct.

class LeetcodeSolution(object):
def twoSum(self, nums, target):
my_map = {}
for i, n in enumerate(nums):
complement = target - n
if my_map.has_key(complement):
return [my_map[complement], i]
my_map
= i