LeetCode | Evaluate Reverse Polish Notation
2017-05-15 18:07
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题目:
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
思路:
逆波兰式计算等式非常容易,利用堆栈即可完成。另外,程序中还需要实现整数与字符串之间的相互转换。代码
“`class Solution {
public:
int evalRPN(vector &tokens) {
int result = 0; int i; stack<int> opd; //存储操作数 int size = tokens.size(); for(i=0;i<size;i++) { if(!strcmp(tokens[i].c_str(),"+")) { int rOpd = opd.top(); //右操作数 opd.pop(); int lOpd = opd.top(); //左操作数 opd.pop(); result = lOpd+rOpd; opd.push(result); } else if(!strcmp(tokens[i].c_str(),"/")) { int rOpd = opd.top(); opd.pop(); int lOpd = opd.top(); opd.pop(); result = lOpd/rOpd; opd.push(result); } else if(!strcmp(tokens[i].c_str(),"-")) { int rOpd = opd.top(); opd.pop(); int lOpd = opd.top(); opd.pop(); result = lOpd-rOpd; opd.push(result); } else if(!strcmp(tokens[i].c_str(),"*")) { int rOpd = opd.top(); opd.pop(); int lOpd = opd.top(); opd.pop(); result = lOpd*rOpd; opd.push(result); } else { opd.push(atoi(tokens[i].c_str())); } } return opd.top(); }
};
思考
1、本题用的是栈2、char *字符串之间的比较用函数strcmp()函数。若用==比较,则默认比较的是地址
char 字符用==符号
3、对于不同的编译器,对string 类型处理不一样。有的编译器会把char 类型数据默认转换为string,有的则不会。所以对于string==char* 的处理需要注意。
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