# twopointer
def twoSum1(self, numbers, target):
l, r = 0, len(numbers)1
while l < r:
s = numbers[l] + numbers[r]
if s == target:
return [l+1, r+1]
elif s < target:
l += 1
else:
r = 1
# dictionary
def twoSum2(self, numbers, target):
dic = {}
for i, num in enumerate(numbers):
if targetnum in dic:
return [dic[targetnum]+1, i+1]
dic[num] = i
# binary search
def twoSum(self, numbers, target):
for i in xrange(len(numbers)):
l, r = i+1, len(numbers)1
tmp = target  numbers[i]
while l <= r:
mid = l + (rl)//2
if numbers[mid] == tmp:
return [i+1, mid+1]
elif numbers[mid] < tmp:
l = mid+1
else:
r = mid1
Python different solutions (twopointer, dictionary, binary search).



Here's a slightly modified version of the Python "dictionary" solution. It sets
enumerate
's second argumentstart=1
and adheres to EAFP instead of LBYL.def twoSum(self, numbers, target): """ two_sum == PEP8 """ seen = {} for i, num in enumerate(numbers, 1): try: return [seen[num], i] except KeyError: seen[target  num] = i

@deliriouslettuce ok,well,you solution is great,but you forget to plus 1 in your result:)

Here's a summary of how long it takes to run for three methods:
 Two pointers: 43 ms
 Dictionary: 46 ms
 Binary Search: 75 ms
It was odd to me that binary search is significantly slower than other two methods; it should be faster since we are using the sorted feature. It turns out that if we don't repeat investigating the elements that have been already investigated, binary search becomes the fastest method. Here's the modified version which takes 35 ms:
def twoSum(self, numbers, target): investigatedSoFar = [] for i in range(len(numbers)): if not numbers[i] in investigatedSoFar: investigatedSoFar.append(numbers[i]) l, r = i + 1, len(numbers)  1 tmp = target  numbers[i] while l <= r: mid = l + (rl) // 2 if numbers[mid] == tmp: return([i + 1, mid + 1]) elif numbers[mid] < tmp: l = mid + 1 else: r = mid  1


@sggkjihua your question is incorrect because the solution is unique per problem description