1+ #include < vector>
2+ #include < algorithm>
3+ #include < iostream>
4+ #include < stack>
5+ using namespace std ;
6+ 7+ class Solution {
8+ public:
9+ int calculate (string s) {
10+ stack<int > nums; // 存储原字符串中的数
11+ stack<char > ops; // 存储原字符串中的符号(含括号和+-)
12+ for (int i = 0 ; i < s.size (); i++)
13+ {
14+ auto c = s[i];
15+ if (c == ' ' continue ; // 过滤空格
16+ if (isdigit (c))
17+ {
18+ int n = 0 ;
19+ int j = i;
20+ while (j < s.size () && isdigit (s[j]))
21+ n = n*10 + (s[j++] - ' 0'
22+ i = j-1 ;
23+ nums.push (n);
24+ }
25+ else if (c == ' (' push (c);
26+ else if (c == ' )'
27+ {
28+ while (ops.top () != ' ('
29+ myEval (nums, ops);
30+ ops.pop ();
31+ }
32+ else
33+ {
34+ if (i == 0 || s[i - 1 ] == ' (' 1 ] == ' +' 1 ] == ' -' // 特殊处理符号和正号
35+ nums.push (0 );
36+ while (!ops.empty () && ops.top () != ' (' /* 由于此题中运算符只有+或-, 故不需要比较不同运算符的优先级了(括号前面已经处理了) */
37+ myEval (nums, ops);
38+ ops.push (c);
39+ }
40+ }
41+ while (!ops.empty ()) myEval (nums, ops);
42+ return nums.top ();
43+ }
44+ void myEval (stack<int >& nums, stack<char >& ops)
45+ {
46+ if (ops.empty ()) return ;
47+ int res;
48+ 49+ auto b = nums.top ();
50+ nums.pop ();
51+ int a = 0 ;
52+ if (nums.size () > 0 )
53+ {
54+ a = nums.top ();
55+ nums.pop ();
56+ }
57+ auto c = ops.top ();
58+ ops.pop ();
59+ 60+ if (c == ' +'
61+ else if (c == ' -'
62+ nums.push (res);
63+ }
64+ };
65+ 66+ // Test
67+ int main ()
68+ {
69+ Solution sol;
70+ // string s = "(1+(4+5+2)-3)+(6+8)"; //"-2+ 1";
71+ string s = " 1-(-2)" // "-2+ 1";
72+ auto res = sol.calculate (s);
73+ cout << res << endl;
74+ 75+ return 0 ;
76+ }
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