282. Expression Add Operators
Read the full problem statement on LeetCode.
Difficulty: hard Acceptance: 41% Topics: Math, String, Backtracking
View full problem on LeetCode Reading material
Reference solution (spoiler · python)
# Time: O(4^n)
# Space: O(n)
class Solution(object):
def addOperators(self, num, target):
"""
:type num: str
:type target: int
:rtype: List[str]
"""
result, expr = [], []
val, i = 0, 0
val_str = ""
while i < len(num):
val = val * 10 + ord(num[i]) - ord('0')
val_str += num[i]
# Avoid "00...".
if str(val) != val_str:
break
expr.append(val_str)
self.addOperatorsDFS(num, target, i + 1, 0, val, expr, result)
expr.pop()
i += 1
return result
def addOperatorsDFS(self, num, target, pos, operand1, operand2, expr, result):
if pos == len(num) and operand1 + operand2 == target:
result.append("".join(expr))
else:
val, i = 0, pos
val_str = ""
while i < len(num):
val = val * 10 + ord(num[i]) - ord('0')
val_str += num[i]
# Avoid "00...".
if str(val) != val_str:
break
# Case '+':
expr.append("+" + val_str)
self.addOperatorsDFS(num, target, i + 1, operand1 + operand2, val, expr, result)
expr.pop()
# Case '-':
expr.append("-" + val_str)
self.addOperatorsDFS(num, target, i + 1, operand1 + operand2, -val, expr, result)
expr.pop()
# Case '*':
expr.append("*" + val_str)
self.addOperatorsDFS(num, target, i + 1, operand1, operand2 * val, expr, result)
expr.pop()
i += 1
Solution from kamyu104/LeetCode-Solutions · MIT