★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★➤微信公众号:山青咏芝(shanqingyongzhi)➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/)➤GitHub地址:https://github.com/strengthen/LeetCode➤原文地址: https://www.cnblogs.com/strengthen/p/10536324.html ➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
Given an expression such as expression = "e + 8 - a + 5" and an evaluation map such as {"e": 1} (given in terms of evalvars = ["e"] and evalints = [1]), return a list of tokens representing the simplified expression, such as ["-1*a","14"]
An expression alternates chunks and symbols, with a space separating each chunk and symbol.A chunk is either an expression in parentheses, a variable, or a non-negative integer.A variable is a string of lowercase letters (not including digits.) Note that variables can be multiple letters, and note that variables never have a leading coefficient or unary operator like "2x" or "-x".
Expressions are evaluated in the usual order: brackets first, then multiplication, then addition and subtraction. For example, expression = "1 + 2 * 3" has an answer of ["7"].
The format of the output is as follows:
For each term of free variables with non-zero coefficient, we write the free variables within a term in sorted order lexicographically. For example, we would never write a term like "b*a*c", only "a*b*c".Terms have degree equal to the number of free variables being multiplied, counting multiplicity. (For example, "a*a*b*c" has degree 4.) We write the largest degree terms of our answer first, breaking ties by lexicographic order ignoring the leading coefficient of the term.The leading coefficient of the term is placed directly to the left with an asterisk separating it from the variables (if they exist.) A leading coefficient of 1 is still printed.An example of a well formatted answer is ["-2*a*a*a", "3*a*a*b", "3*b*b", "4*a", "5*c", "-6"] Terms (including constant terms) with coefficient 0 are not included. For example, an expression of "0" has an output of [].
Examples:
Input: expression = "e + 8 - a + 5", evalvars = ["e"], evalints = [1]
Output: ["-1*a","14"]
Input: expression = "e - 8 + temperature - pressure",
evalvars = ["e", "temperature"], evalints = [1, 12]
Output: ["-1*pressure","5"]
Input: expression = "(e + 8) * (e - 8)", evalvars = [], evalints = []
Output: ["1*e*e","-64"]
Input: expression = "7 - 7", evalvars = [], evalints = []
Output: []
Input: expression = "a * b * c + b * a * c * 4", evalvars = [], evalints = []
Output: ["5*a*b*c"]
Input: expression = "((a - b) * (b - c) + (c - a)) * ((a - b) + (b - c) * (c - a))",
evalvars = [], evalints = []
Output: ["-1*a*a*b*b","2*a*a*b*c","-1*a*a*c*c","1*a*b*b*b","-1*a*b*b*c","-1*a*b*c*c","1*a*c*c*c","-1*b*b*b*c","2*b*b*c*c","-1*b*c*c*c","2*a*a*b","-2*a*a*c","-2*a*b*b","2*a*c*c","1*b*b*b","-1*b*b*c","1*b*c*c","-1*c*c*c","-1*a*a","1*a*b","1*a*c","-1*b*c"]
Note:
expression will have length in range [1, 250].evalvars, evalints will have equal lengths in range [0, 100].
给定一个表达式 expression 如 expression = "e + 8 - a + 5" 和一个求值映射,如 {"e": 1}(给定的形式为 evalvars = ["e"] 和 evalints = [1]),返回表示简化表达式的标记列表,例如 ["-1*a","14"]
表达式交替使用块和符号,每个块和符号之间有一个空格。块要么是括号中的表达式,要么是变量,要么是非负整数。块是括号中的表达式,变量或非负整数。变量是一个由小写字母组成的字符串(不包括数字)。请注意,变量可以是多个字母,并注意变量从不具有像 "2x" 或 "-x" 这样的前导系数或一元运算符 。
表达式按通常顺序进行求值:先是括号,然后求乘法,再计算加法和减法。例如,expression = "1 + 2 * 3" 的答案是 ["7"]。
输出格式如下:
对于系数非零的每个自变量项,我们按字典排序的顺序将自变量写在一个项中。例如,我们永远不会写像 “b*a*c” 这样的项,只写 “a*b*c”。项的次数等于被乘的自变量的数目,并计算重复项。(例如,"a*a*b*c" 的次数为 4。)。我们先写出答案的最大次数项,用字典顺序打破关系,此时忽略词的前导系数。项的前导系数直接放在左边,用星号将它与变量分隔开(如果存在的话)。前导系数 1 仍然要打印出来。格式良好的一个示例答案是 ["-2*a*a*a", "3*a*a*b", "3*b*b", "4*a", "5*c", "-6"] 。系数为 0 的项(包括常数项)不包括在内。例如,“0” 的表达式输出为 []。
示例:
输入:expression = "e + 8 - a + 5", evalvars = ["e"], evalints = [1]
输出:["-1*a","14"]
输入:expression = "e - 8 + temperature - pressure",
evalvars = ["e", "temperature"], evalints = [1, 12]
输出:["-1*pressure","5"]
输入:expression = "(e + 8) * (e - 8)", evalvars = [], evalints = []
输出:["1*e*e","-64"]
输入:expression = "7 - 7", evalvars = [], evalints = []
输出:[]
输入:expression = "a * b * c + b * a * c * 4", evalvars = [], evalints = []
输出:["5*a*b*c"]
输入:expression = "((a - b) * (b - c) + (c - a)) * ((a - b) + (b - c) * (c - a))",
evalvars = [], evalints = []
输出:["-1*a*a*b*b","2*a*a*b*c","-1*a*a*c*c","1*a*b*b*b","-1*a*b*b*c","-1*a*b*c*c","1*a*c*c*c","-1*b*b*b*c","2*b*b*c*c","-1*b*c*c*c","2*a*a*b","-2*a*a*c","-2*a*b*b","2*a*c*c","1*b*b*b","-1*b*b*c","1*b*c*c","-1*c*c*c","-1*a*a","1*a*b","1*a*c","-1*b*c"]
提示:
expression 的长度在 [1, 250] 范围内。evalvars, evalints 在范围 [0, 100] 内,且长度相同。
同样的测试用例。
使用【测试用例】单独执行,执行正确。
input
"(e + 8) * (e - 8)"
[]
[]
Output
["1*e*e","-64"]
Expected
["1*e*e","-64"]
但是提交时,提示解答错误。问题不大。
Input
"(e + 8) * (e - 8)"
[]
[]
Output
["-63"]
Expected
["1*e*e","-64"]
1 class Solution {
2 //evaluation map
3 static var map:[String:Int] =
[String:Int]()
4
5 func basicCalculatorIV(_ expression: String, _ evalvars: [String], _ evalints: [Int]) ->
[String] {
6 for i
in 0..<
evalvars.count
7 {
8 Solution.map[evalvars[i]] =
evalints[i]
9 }
10 var i:Int =
0
11 var l:Int =
expression.count
12 var stack:[Expression] =
[Expression]()
13 var priStack:[Int] =
[Int]()
14 var zero:Expression = Expression(
0)
15 stack.append(zero)
16 priStack.append(
0)
17 var pri:Int =
0
18
19 while (i <
l)
20 {
21 var ch:Character =
expression[i]
22 if ch.isDigit()
23 {
24 var num:Int =
0
25 while (i < l &&
expression[i].isDigit())
26 {
27 num = num *
10 + (expression[i].ascii -
48)
28 i +=
1
29 }
30 stack.append(Expression(num))
31 continue
32 }
33 if ch.isLetter()
34 {
35 var s:String =
String()
36 while (i < l &&
expression[i].isLetter())
37 {
38 s.append(expression[i])
39 i +=
1
40 }
41 stack.append(Expression(s))
42 continue
43 }
44 if ch ==
"(" {pri +=
2}
45 if ch ==
")" {pri -=
2}
46 if ch ==
"+" || ch ==
"-" || ch ==
"*"
47 {
48 var nowPri:Int =
pri
49 if ch ==
"*" {nowPri +=
1}
50 while (!priStack.isEmpty && nowPri <= priStack.last!
)
51 {
52 var now:Expression =
stack.removeLast()
53 var last:Expression =
stack.removeLast()
54 priStack.removeLast()
55 stack.append(last.cal(now))
56 }
57 stack[stack.count -
1].oper =
ch
58 priStack.append(nowPri)
59 }
60 i +=
1
61 }
62
63 while (stack.count >
1)
64 {
65 var now:Expression =
stack.removeLast()
66 var last:Expression =
stack.removeLast()
67 stack.append(last.cal(now))
68 }
69
70 return stack.last!
.toList()
71 }
72 }
73
74 class Term
75 {
76 //the parameter of this term
77 var para:Int =
1
78 //each factor (e.a. a*b*b*c->{a,b,b,c})
79 var arr:[String] =
[String]()
80
81 init(_ x:Int)
82 {
83 self.para =
x
84 }
85
86 init(_ s:String)
87 {
88 if Solution.map[s] ==
nil
89 {
90 arr.append(s)
91 }
92 else
93 {
94 para = Solution.map[s]!
95 }
96 }
97
98 init(_ that:Term)
99 {
100 self.para =
that.para
101 self.arr =
that.arr
102 }
103
104 func toString() ->
String
105 {
106 if para ==
0 {
return String()}
107 var ans:String =
String()
108 for s
in arr
109 {
110 ans += (
"*" +
s)
111 }
112 return String(para) +
ans
113 }
114
115 func equals(_ that:Term) ->
Bool
116 {
117 if self.arr.count !=
that.arr.count
118 {
119 return false
120 }
121 for i
in 0..<
self.arr.count
122 {
123 if self.arr[i] !=
that.arr[i]
124 {
125 return false
126 }
127 }
128 return true
129 }
130
131 func compareTo(_ that:Term) ->
Int
132 {
133 if self.arr.count > that.arr.count {
return -
1}
134 if self.arr.count < that.arr.count {
return 1}
135 for i
in 0..<
self.arr.count
136 {
137 var x:Int =
0
138 if self.arr[i] >
that.arr[i]
139 {
140 x =
1
141 }
142 else if self.arr[i] <
that.arr[i]
143 {
144 x = -
1
145 }
146 else
147 {
148 x =
0
149 }
150 if x !=
0 {
return x}
151 }
152 return 0
153 }
154
155 func times(_ that:Term) ->
Term
156 {
157 var pro = Term(self.para *
that.para)
158 for s
in self.arr
159 {
160 pro.arr.append(String(s))
161 }
162 for s
in that.arr
163 {
164 pro.arr.append(String(s))
165 }
166 pro.arr = pro.arr.sorted(by:>
)
167 return pro
168 }
169 }
170
171 class Expression
172 {
173 //Term List
174 var list:[Term] =
[Term]()
175 var oper:Character =
"+"
176
177 init(_ x:Int)
178 {
179 self.list.append(Term(x))
180 }
181
182 init(_ s:String)
183 {
184 self.list.append(Term(s))
185 }
186
187 init(_ l:[Term])
188 {
189 self.list =
l
190 }
191
192 func times(_ that:Expression) ->
Expression
193 {
194 var c:[Term] =
[Term]()
195 for t1
in self.list
196 {
197 for t2
in that.list
198 {
199 c.append(t1.times(t2))
200 }
201 }
202 c =
combine(c)
203 return Expression(c)
204 }
205
206 func plus(_ that:Expression,_ sgn:Int) ->
Expression
207 {
208 var c:[Term] =
[Term]()
209 for t
in self.list
210 {
211 c.append(Term(t))
212 }
213 for t
in that.list
214 {
215 var t2:Term =
Term(t)
216 t2.para = t2.para *
sgn
217 c.append(t2)
218 }
219 c =
combine(c)
220 return Expression(c)
221 }
222
223 func cal(_ that:Expression) ->
Expression
224 {
225 if oper ==
"+"
226 {
227 return plus(that,
1)
228 }
229 if oper ==
"-"
230 {
231 return plus(that,-
1)
232 }
233 return times(that)
234 }
235
236 func toList() ->
[String]
237 {
238 var ans:[String] =
[String]()
239 for t
in list
240 {
241 var s:String =
t.toString()
242 if s.count >
0 {ans.append(s)}
243 }
244 return ans
245 }
246
247 func combine(_ a:[Term]) ->
[Term]
248 {
249 var a =
a
250 a.sort(by:{(t1:Term,t2:Term) -> Bool
in
251 return t1.compareTo(t2) <
0})
252 var c:[Term] =
[Term]()
253 for t
in a
254 {
255 if c.count !=
0 && t.equals(c.last!
)
256 {
257 c[c.count -
1].para +=
t.para
258 }
259 else
260 {
261 c.append(Term(t));
262 }
263 }
264 return c
265 }
266 }
267
268 //String扩展
269 extension String {
270 //subscript函数可以检索数组中的值
271 //直接按照索引方式截取指定索引的字符
272 subscript (_ i: Int) ->
Character {
273 //读取字符
274 get {
return self[index(startIndex, offsetBy: i)]}
275 }
276 }
277
278 //Character扩展
279 extension Character
280 {
281 //Character转ASCII整数值(定义小写为整数值)
282 var ascii: Int {
283 get {
284 return Int(self.unicodeScalars.first?.value ??
0)
285 }
286 }
287
288 func isDigit() ->
Bool
289 {
290 if self.ascii >
47 && self.ascii <
58
291 {
292 return true
293 }
294 return false
295 }
296
297 func isLetter() ->
Bool
298 {
299 if (self.ascii >
64 && self.ascii <
91) || self.ascii >
96 && self.ascii <
123
300 {
301 return true
302 }
303 return false
304 }
305 }
转载于:https://www.cnblogs.com/strengthen/p/10536324.html