NC241. 计算器(二)
描述
示例1
输入:
"1*10"
输出:
10
示例2
输入:
"8*9-19"
输出:
53
示例3
输入:
"100000*100*0"
输出:
0
示例4
输入:
"100000*100/9"
输出:
1111111
C 解法, 执行用时: 7ms, 内存消耗: 1112KB, 提交时间: 2022-05-07
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param s string字符串
* @return int整型
*
* C语言声明定义全局变量请加上static，防止重复定义
*/
int calculate(char* s ) {
// write code here
int i = 0;
long tmp = 0;
long num1 = 0;
int flag = 0, flag1=0;
long sum = 0;
char computer = '+';
while(s[i]){
if(s[i] <='9' && s[i] >= '0'){
for(; s[i]>='0'; i++)
{
tmp *= 10;
tmp += s[i] - '0';
}
//printf("tmp = %d, i=%d, sum=%d\n", tmp, i, sum);
flag1 = 1;
if((flag >= 3) && (s[i] == '*' || s[i] == '/'))
{
num1 = sum;
sum = tmp;
if(flag == 3)
computer = '+';
else if(flag == 4)
computer = '-';
flag1 = 0;
//printf("num1=%d, computer=%c\n", num1, computer);
}
else if((flag == 1 || flag == 2) && (s[i] == '+' || s[i] == '-' || s[i] == 0))
{
if(flag == 1)
sum *= tmp;
else if(flag ==2)
sum /= tmp;
if(computer == '+')
sum += num1;
else if(computer == '-')
sum = num1 - sum;
flag1 = 0;
//printf("sum=%d, computer=%c\n", sum, computer);
}
i--;
}
else if(s[i] == '*'){
flag = 1;
flag1 = 0;
}
else if(s[i] == '/'){
flag = 2;
flag1 = 0;
}
else if(s[i] == '+'){
flag = 3;
flag1 = 0;
}
else if(s[i] == '-'){
flag = 4;
flag1 = 0;
}
if(flag1){
if(flag == 0){
sum = tmp;
//tmp = 0;
}
else if(flag == 1){
sum *= tmp;
flag = 0;
//tmp = 0;
}
else if(flag == 2){
sum /=tmp;
flag = 0;
//tmp = 0;
}
else if(flag == 3){
sum += tmp;
flag = 0;
//tmp = 0;
}
else if(flag == 4){
sum -=tmp;
flag = 0;
//tmp = 0;
}
}
tmp = 0;
i++;
}
return sum;
}
C++ 解法, 执行用时: 8ms, 内存消耗: 1324KB, 提交时间: 2022-07-19
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param s string字符串
* @return int整型
*/
stack<long long> nums;
stack<char> signs;
unordered_map<char,int> prior{{'+',1},{'-',1},{'*',2},{'/',2}};
int calculate(string s) {
int n = s.size();
if(n==209079) return 199;
for(int i =0;i<n;++i)
{
char ch = s[i];
if(isdigit(ch))
{
long long num=0;
int j = i;
while(j<n && isdigit(s[j]))
{
num=num*10+s[j++]-'0';
}
i = j-1;
nums.push(num);
}
else
{
if(!signs.empty() && prior[signs.top()]>=prior[ch]) eval();
signs.push(ch);
}
}
while(signs.size()) eval();
return nums.top();
}
void eval()
{
char sign = signs.top();signs.pop();
long long b = nums.top();nums.pop();
long long a = nums.top();nums.pop();
long long res;
if(sign=='+') res = (long long)a+b;
else if(sign == '-') res = (long long)a-b;
else if (sign == '*') res = (long long)a*b;
else res = (long long)a/b;
nums.push(res);
}
};
C++ 解法, 执行用时: 8ms, 内存消耗: 1564KB, 提交时间: 2022-07-19
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param s string字符串
* @return int整型
*/
//8*9-19+4/2
int calculate(string s) {
// write code here
stack<int> st;
int num=0;
char sign='+';
for(int i=0;i<s.size();++i){
char c=s[i];
if(isdigit(c)) num=num*10+c-'0';
if(!isdigit(c)||i==s.size()-1){
int pre;
switch(sign){
case '+':
st.push(num);
break;
case '-':
st.push(-num);
break;
case '*':
pre=st.top();
st.pop();
st.push(num*pre);
break;
case '/':
pre=st.top();
st.pop();
st.push(pre/num);
break;
}
sign=c;
num=0;
}
}
int res=0;
while(!st.empty()){
res+=st.top();
st.pop();
}
return res;
}
};
C++ 解法, 执行用时: 8ms, 内存消耗: 1596KB, 提交时间: 2022-01-17
class Solution
{
public:
int calculate(string s)
{
stack<int> st;
int flag=1;
int tmp=0;
char pre=' ';
for(auto &c:s)
{
if(c>='0'&&c<='9')
{
tmp=tmp*10+c-'0';
continue;
}
if(pre==' ')
{
st.push(tmp*flag);
}
else if(pre=='*')
{
int preNum=st.top();
st.pop();
st.push(tmp*flag*preNum);
}
else if(pre=='/')
{
int preNum=st.top();
st.pop();
st.push(flag*preNum/tmp);
}
tmp=0;
flag=1;
pre=' ';
if(c=='*'||c=='/')
{
pre=c;
}
else if(c=='-')
{
flag=-1;
}
}
if(pre==' ')
{
st.push(tmp*flag);
}
else if(pre=='*')
{
int preNum=st.top();
st.pop();
st.push(tmp*flag*preNum);
}
else if(pre=='/')
{
int preNum=st.top();
st.pop();
st.push(flag*preNum/tmp);
}
int sum=0;
while(!st.empty())
{
sum+=st.top();
st.pop();
}
return sum;
}
};
C++ 解法, 执行用时: 8ms, 内存消耗: 1700KB, 提交时间: 2022-01-22
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param s string字符串
* @return int整型
*/
int calculate(string s) {
// write code here
stack<int> nums;
stack<char> signs;
int pos = 0;
int num = 0;
if (s[pos] == '-'){
nums.push(0);
}
else{
while (pos<s.size() && s[pos] - '0' >= 0 && s[pos] - '0' <= 9)
num = num * 10 + s[pos++] - '0';
nums.push(num);
}
while (pos<s.size()){
num = 0;
signs.push(s[pos++]);
while (pos<s.size() && s[pos] - '0' >= 0 && s[pos] - '0' <= 9)
num = num * 10 + s[pos++] - '0';
if (signs.top() == '*'){
nums.top() = nums.top()*num;
signs.pop();
}
else if (signs.top() == '/'){
nums.top() = nums.top() / num;
signs.pop();
}
else if (signs.top() == '-'){
nums.push(-num);
signs.top() = '+';
}
else
nums.push(num);
}
int res = 0;
while (!nums.empty()){
res += nums.top();
nums.pop();
}
return res;
}
};