Leetcode 227. Basic Calculator II
Leetcode 227. Basic Calculator II
triplebee
发布于 2018-01-12 14:39:58
发布于 2018-01-12 14:39:58
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.
计算字符串表达式的值。
模拟一下，遇到+或-就连数值一起放到一个容器中，遇到*或/则优先处理，最后将容器中的值都加起来。
class Solution {
public:
int calculate(string s) {
stack<int> ss;
int now = 0, flag = 1, temp;
s += "#";
bool flagd = false, flagm = false;
for(int i = 0; i < s.size(); i ++)
{
if(s[i] == ' ') continue;
if(s[i] >= '0' && s[i] <= '9') now = now * 10 + s[i] - '0';
else if(flagm)
{
i--;
flagm = false;
now *= temp;
}
else if(flagd)
{
i--;
flagd = false;
now = temp / now;
}
else if(s[i] == '#') continue;
else if(s[i] == '+')
{
ss.push(flag * now);
flag = 1;
now = 0;
}
else if(s[i] == '-')
{
ss.push(flag * now);
flag = -1;
now = 0;
}
else
{
temp = now;
now = 0;
if(s[i] == '*') flagm = true;
else flagd = true;
}
}
now *= flag;
while(!ss.empty())
{
now += ss.top();
ss.pop();
}
return now;
}
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原始发表：2017-03-30 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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