难度: Medium
原题连接
内容描述
Solve a given equation and return the value of x in the form of string "x=#value". The equation contains only '+', '-' operation, the variable x and its coefficient.
If there is no solution for the equation, return "No solution".
If there are infinite solutions for the equation, return "Infinite solutions".
If there is exactly one solution for the equation, we ensure that the value of x is an integer.
Example 1:
Input: "x+5-3+x=6+x-2"
Output: "x=2"
Example 2:
Input: "x=x"
Output: "Infinite solutions"
Example 3:
Input: "2x=x"
Output: "x=0"
Example 4:
Input: "2x+3x-6x=x+2"
Output: "x=-1"
Example 5:
Input: "x=x+2"
Output: "No solution"
思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******
跟calculator一样，beats 32.13%
class Solution:
def solveEquation(self, equation):
"""
:type equation: str
:rtype: str
"""
def helper(s):
sign, n = 1, len(s)
# i, coef, const stand for current index, and accumulative 'x' coefficient and constant
i = coef = const = 0
while i < n:
if s[i] == '+':
sign = 1
elif s[i] == '-':
sign = -1
elif s[i].isdigit():
j = i
while j < n and s[j].isdigit():
j += 1
tmp = int(s[i:j])
if j < n and s[j] == 'x':
coef += tmp * sign
j += 1
else:
const += tmp * sign
i = j-1
else:
coef += 1 * sign
i += 1
return coef, const
left, right = equation.split('=')
k1, b1 = helper(left)
k2, b2 = helper(right)
# k1x + b1 = k2x + b2
if k1 != k2 and b1 != b2:
res = 'x=' + str((b2 - b1) // (k1 - k2))
elif k1 == k2 and b1 == b2:
res = 'Infinite solutions'
elif k1 == k2 and b1 != b2:
res = 'No solution'
elif k1 != k2 and b1 == b2:
res = 'x=0'
return res