[LeetCode] 344. Reverse String
Contents
Reverse String
Link to original Problem on LeetCode
Write a function that reverses a string. The input string is given as an array of characters char[].
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
You may assume all the characters consist of printable ascii characters.
Example 1:
Input: [“h”,“e”,“l”,“l”,“o”] Output: [“o”,“l”,“l”,“e”,“h”]
Example 2:
Input: [“H”,“a”,“n”,“n”,“a”,“h”] Output: [“h”,“a”,“n”,“n”,“a”,“H”]
Company:
Solution (Python Way):
class Solution(object):
def reverseString(self, s):
"""
:type s: List[str]
:rtype: None Do not return anything, modify s in-place instead.
"""
# Using [::-1] to reverse the string
s[:] = s[::-1]
Solution (Iterative):
class Solution(object):
def reverseString(self, s):
"""
:type s: List[str]
:rtype: None Do not return anything, modify s in-place instead.
"""
# Keep two reference to the first and the last element of the list, say left and right respectively
# Now until left is greater than right, swap left and right element and increament left and decreament right
left, right = 0, len(s) - 1
while left <= right:
s[left], s[right] = s[right], s[left]
left += 1
right -= 1
Solution (Recursive):
class Solution(object):
def reverseString(self, s):
"""
:type s: List[str]
:rtype: None Do not return anything, modify s in-place instead.
"""
# Logic is same as the one in iterative solution
if not len(s):
return []
self.reverseStringHelper(0, len(s) - 1, s)
def reverseStringHelper(self, left, right, s):
if left >= right:
return
s[left], s[right] = s[right], s[left]
self.reverseStringHelper(left + 1, right - 1, s)