标签：
Basic Calculator II
Implement a basic calculator to evaluate a simple expression string.
The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.
You may assume that the given expression is always valid.
Some examples:
"3+2*2" = 7 " 3/2 " = 1 " 3+5 / 2 " = 5
Note: Do not use the eval built-in library function.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
与Basic Calculator对照，
乘除法优先级高，因此一旦出现就立即计算，
到最后栈中剩下的就是顺序执行的加减计算。
先将栈逆序，再出栈计算。注意，此时的操作数先后顺序反过来。
class Solution { public: int calculate(string s) { stack<int> num; stack<char> op; int i = 0; while(i < s.size()) { while(i < s.size() && s[i] == ‘ ‘) i ++; if(i == s.size()) break; if(s[i] == ‘+‘ || s[i] == ‘-‘ || s[i] == ‘*‘ || s[i] == ‘/‘) { op.push(s[i]); i ++; } else { int n = 0; while(i < s.size() && s[i] >= ‘0‘ && s[i] <= ‘9‘) { n = n * 10 + (s[i]-‘0‘); i ++; } num.push(n); if(!op.empty() && (op.top() == ‘*‘ || op.top() == ‘/‘)) { int n2 = num.top(); num.pop(); int n1 = num.top(); num.pop(); if(op.top() == ‘*‘) num.push(n1 * n2); else num.push(n1 / n2); op.pop(); } } } // ‘+‘/‘-‘ in order if(!op.empty()) { // reverse num and op stack<int> num2; while(!num.empty()) { num2.push(num.top()); num.pop(); } num = num2; stack<char> op2; while(!op.empty()) { op2.push(op.top()); op.pop(); } op = op2; while(!op.empty()) { // pay attention to the operands order! int n1 = num.top(); num.pop(); int n2 = num.top(); num.pop(); if(op.top() == ‘+‘) num.push(n1 + n2); else num.push(n1 - n2); op.pop(); } } return num.top(); } };
【LeetCode】227. Basic Calculator II
标签：
原文地址：http://www.cnblogs.com/ganganloveu/p/4640966.html
