## Leetcode 20190301

https://leetcode.com/explore/featured/card/top-interview-questions-easy/92/array/646/

## Strings

### Implement strStr()

忘记用 Python 的 [x:y] 来一段段匹配了：

```
if haystack == needle or not needle:
return 0
for i, j in enumerate(haystack):
if i > len(haystack) - len(needle) + 1:
break
if needle and haystack[i] == needle[0]:
if i + len(needle) <= len(haystack):
notMatch = False
for l, m in enumerate(needle):
if haystack[i + l] != needle[l]:
notMatch = True
break
if not notMatch:
return i
return -1
```

Runtime: 44 ms, faster than 55.70% of Python3 online submissions for Implement strStr().

Memory Usage: 13.4 MB, less than 5.13% of Python3 online submissions for Implement strStr().

### 1. Two Sum

https://leetcode.com/problems/two-sum/

```
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
if nums[i] + nums[j] == target:
return [i, j]
```

Runtime: 5424 ms, faster than 12.11% of Python3 online submissions for Two Sum.

Memory Usage: 13.6 MB, less than 42.31% of Python3 online submissions for Two Sum.

### 2. Add Two Numbers

https://leetcode.com/problems/add-two-numbers/

```
class Solution:
def addTwoNumbers(self, l1, l2):
head = None
last = None
while l1 is not None or l2 is not None:
i = (0 if l1 is None else l1.val) + (0 if l2 is None else l2.val)
if l1 is not None:
l1 = l1.next
if l2 is not None:
l2 = l2.next
if last is None:
head = ListNode(i)
last = head
else:
if last.next is None:
last.next = ListNode(i)
else:
last.next.val += i
last = last.next
if last.val >= 10:
last.val -= 10
last.next = ListNode(1)
return head
```

Runtime: 112 ms, faster than 61.12% of Python3 online submissions for Add Two Numbers.

Memory Usage: 13.4 MB, less than 5.21% of Python3 online submissions for Add Two Numbers.

### 3. Longest Substring Without Repeating Characters

https://leetcode.com/problems/longest-substring-without-repeating-characters/

```
class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
count_max = 0
for i in range(len(s)):
k = []
count = 0
for j in range(i, len(s)):
if s[j] in k:
break
else:
k.append(s[j])
count += 1
if count > count_max:
count_max = count
return count_max
```

Runtime: 1796 ms, faster than 5.02% of Python3 online submissions for Longest Substring Without Repeating Characters.

Memory Usage: 13.4 MB, less than 5.05% of Python3 online submissions for Longest Substring Without Repeating Characters.