目錄
227. Basic Calculator II
Medium
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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.
Example 1:
Input: "3+2*2"
Output: 7
Example 2:
Input: " 3/2 "
Output: 1
Example 3:
Input: " 3+5 / 2 "
Output: 5
Note:
- You may assume that the given expression is always valid.
- Do not use the
evalbuilt-in library function.
class Solution {
public:
int calculate(string s) {
char sign='+';
int num=0;
stack<int> st;
for(int i=0;i<s.size();i++){
if(s[i]>='0' && s[i]<='9') num=num*10+(s[i]-'0');
if(s[i]=='+' || s[i]=='-' || s[i]=='*' || s[i]=='/' || i==s.size()-1){
if(sign=='+') st.push(num);
else if(sign=='-') st.push((-1)*num);
else if(sign=='*'){
int a=st.top();
st.pop();
st.push(a*num);
}else if(sign=='/'){
int a=st.top();
st.pop();
st.push(a/num);
}
num=0;
sign=s[i];
}
}
int res=0;
while(!st.empty()){
int a=st.top();
st.pop();
if(!st.empty()){
int b=st.top();
st.pop();
st.push(a+b);
}else res=a;
}
return res;
}
};
- 遇到 ‘ + | - | * | / ’ 或倒數最後一個字符：
- sign= ‘ + ’ ：st.push(num)
- sign= ‘ - ’ ：st.push((-1)*num)
- sign= ‘ * ’ ：top=st.top();st.pop();st.push(top*num);
- sign= ‘ / ’ ：top=st.top();st.pop();st.push(top/num);
- sign=s[i],num=0
224. Basic Calculator
Hard
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Implement a basic calculator to evaluate a simple expression string.
The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces .
Example 1:
Input: "1 + 1"
Output: 2
Example 2:
Input: " 2-1 + 2 "
Output: 3
Example 3:
Input: "(1+(4+5+2)-3)+(6+8)"
Output: 23
Note:
- You may assume that the given expression is always valid.
- Do not use the
evalbuilt-in library function.
class Solution {
public:
int calculate(string s) {
int res = 0, num = 0, sign = 1, n = s.size();
stack<int> st;
for (int i = 0; i < n;i++){
char c = s[i];
if(c>='0' && c<='9') num=num*10+(c-'0');
if(c=='+' || c=='-'){
res += sign * num;
num = 0;
sign = (c == '+') ? 1 : -1;
}else if(c=='('){
st.push(res);
st.push(sign);
res = 0;
sign = 1;
}else if(c==')'){
res += sign * num;
num = 0;
res *= st.top();//正還是負
st.pop();
res += st.top();
st.pop();
}
}
res += sign * num;
return res;
}
};
- 只有遇到 ‘（ ’ 才入棧；
- 遇到 ‘（ ’ ：push(res),push(sign),res=0,sign=1
- 遇到 ‘ ）’ ：res+=num*sign,res*=st.top();st.pop();res+=st.top();st.pop();num=0;
- 遇到 ‘ + ’ ：res+=num*sign,num=0,sign=1
- 遇到 ‘ - ’ ：res+=num*sign,num=0,sign=-1
282. Expression Add Operators
Hard
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Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +, -, or * between the digits so they evaluate to the target value.
Example 1:
Input: num = "123", target = 6
Output: ["1+2+3", "1*2*3"]
Example 2:
Input: num = "232", target = 8
Output: ["2*3+2", "2+3*2"]
Example 3:
Input: num = "105", target = 5
Output: ["1*0+5","10-5"]
Example 4:
Input: num = "00", target = 0
Output: ["0+0", "0-0", "0*0"]
Example 5:
Input: num = "3456237490", target = 9191
Output: []
#include"pch.h"
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<stack>
using namespace std;
/*282. Expression Add Operators
1.DFS，計算出每個s的最終值，超時*/
long long getNum(string s) {
stack<long long> st;
long long num = 0;
char flag = '+';
for (int i = 0; i < s.size(); i++) {
if (s[i] >= '0' && s[i] <= '9') num = num * 10 + (s[i] - '0');
if(s[i]=='+' || s[i]=='-' || s[i]=='*' || i==s.size()-1){
if (flag == '*') {
long long a = st.top();
st.pop();
st.push(a*num);
}
else if (flag == '+') st.push(num);
else if (flag == '-') st.push(-num);
flag = s[i];
num = 0;
}
}
long long res = 0;
while (!st.empty()) {
long long a = st.top();
st.pop();
if (!st.empty()) {
long long b = st.top();
st.pop();
st.push(a + b);
}
else res = a;
}
return res;
}
void DFS(string num, int start, int target, vector<string> &out, vector<string> &res) {
if (start >= num.size()) {
string s = "";
for (auto a : out) s += a;
if (getNum(s) == target) res.push_back(s);
return;
}
int i = start;
int right = num.size() - i;
if (num[i] == '0') right = 1;
for (int j = 1; j <= right; j++) {
string s = num.substr(i, j);
out.push_back(s);
if (i + j != num.size()) {
out.push_back("+");
DFS(num, i + j, target, out, res);
out.pop_back();
out.push_back("-");
DFS(num, i + j, target, out, res);
out.pop_back();
out.push_back("*");
DFS(num, i + j, target, out, res);
out.pop_back();
out.pop_back();
}
else {
DFS(num, i + j, target, out, res);
out.pop_back();
}
}
return ;
}
vector<string> addOperators1(string num, int target) {
vector<string> res,out;
DFS(num, 0, target,out,res);
return res;
}
/*2.兩個變量diff和curNum，
一個用來記錄將要變化的值，另一個是當前運算後的值，
而且它們都需要用 long 型的，因爲字符串轉爲int型很容易溢出，
對於加法，diff就是即將要加上的值，
對於減法，diff就是即將要減去的值的相反數，
對於乘法，diff就是上一個diff乘以即將要乘上的數字，
例如，2+3*2，將運算到*2時，上次循環的curNum=5，diff=3，
算完*2後，diff變爲6，並且把之前的+3操作去掉，再加上新的diff，
即（5-3）+6-8.*/
vector<string> addOperators(string num, int target) {
vector<string> res;
helper(num, target, 0, 0, "", res);
return res;
}
void helper(string num, int target, long diff, long curNum, string out, vector<string>& res) {
if (num.size() == 0 && curNum == target) {
res.push_back(out); return;
}
for (int i = 1; i <= num.size(); ++i) {
string cur = num.substr(0, i);
if (cur.size() > 1 && cur[0] == '0') return;
string next = num.substr(i);
if (out.size() > 0) {
helper(next, target, stoll(cur), curNum + stoll(cur), out + "+" + cur, res);
helper(next, target, -stoll(cur), curNum - stoll(cur), out + "-" + cur, res);
helper(next, target, diff * stoll(cur), (curNum - diff) + diff * stoll(cur), out + "*" + cur, res);
}
else {
helper(next, target, stoll(cur), stoll(cur), cur, res);
}
}
}
int main() {
vector<string> res = addOperators("3456237490",9191);
for (auto a : res) cout << a << " ";
return 0;
}