LeetCode_Evaluate Reverse Polish Notation
2014-05-05 06:34
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Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
题目很简单,就是给出一个公式的波兰表示法,进一步求公式的解,所有解都以整数表示。
这道题真的很简单,选好数据结构即可,对于一个具有两个操作数的公式的计算,只需确定左操作数、右操作数和运算符即可,由此选定栈作为基本数据结构。代码如下:
Valid operators are +, -, *, /. Each operand may be an integer or another expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
题目很简单,就是给出一个公式的波兰表示法,进一步求公式的解,所有解都以整数表示。
这道题真的很简单,选好数据结构即可,对于一个具有两个操作数的公式的计算,只需确定左操作数、右操作数和运算符即可,由此选定栈作为基本数据结构。代码如下:
class Solution { public: int evalRPN(vector<string> &tokens) { stack<int> operNumStack; vector <string>::iterator iter; int right; int left; int result; for (iter=tokens.begin();iter<tokens.end();iter++) { string currentToken=*iter; //加法 if (currentToken.compare("+")==0) { left=operNumStack.top(); operNumStack.pop(); right=operNumStack.top(); operNumStack.pop(); result=right+left; operNumStack.push(result); } else { //减法 if (currentToken.compare("-")==0) { right=operNumStack.top(); operNumStack.pop(); left=operNumStack.top(); operNumStack.pop(); result=left-right; operNumStack.push(result); } else { //乘法 if (currentToken.compare("*")==0) { right=operNumStack.top(); operNumStack.pop(); left=operNumStack.top(); operNumStack.pop(); result=right*left; operNumStack.push(result); } else { if (currentToken.compare("/")==0) { right=operNumStack.top(); operNumStack.pop(); left=operNumStack.top(); operNumStack.pop(); result=left/right; operNumStack.push(result); } else { int currentNum=atoi(currentToken.c_str()); operNumStack.push(currentNum); } } } } } result=operNumStack.top(); operNumStack.pop(); /*cout<<result<<endl;*/ return result; } };
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