数值表达式的计算-leetcode basic calculator

曾经终败给现在 2021-10-29 06:54 348阅读 0赞

1 public class Solution {
     2     
     3     int priority(char op){
     4         if(op == '+' || op == '-') return 1;
     5         else return 2;
     6     }
     7     boolean isNum(String s){
     8         return Character.isDigit(s.charAt(0));
     9     }
    10     boolean isOp(char ch){
    11         return ch == '+' || ch == '-' || ch == '*' || ch == '/';
    12     }
    13     long cal(long num1, char op, long num2){
    14         if(op == '-') return num1 - num2;
    15         else if(op == '+') return num1 + num2;
    16         else if(op == '*') return num1 * num2;
    17         else return num1 / num2;
    18     }
    19     void popOpAndTwoNums(Stack<Character> opStack, Stack<Long> numStack){
    20         char op = opStack.pop();
    21         long num2 = numStack.pop(), num1 = numStack.pop();
    22         long num = cal(num1, op, num2);
    23         numStack.push(num);
    24     }
    25     public ArrayList<String> split(String s){
    26         ArrayList<String> array = new ArrayList<String>();
    27         int i = 0;
    28         while(i < s.length()){
    29             if(s.charAt(i) == '(' || s.charAt(i) == ')' || isOp(s.charAt(i))){
    30                 array.add(s.substring(i, i + 1));
    31                 i++;
    32             }else if(s.charAt(i) == ' '){
    33                 i++;
    34                 continue ;
    35             }else{
    36                 int j = i;
    37                 while(j < s.length() && Character.isDigit(s.charAt(j))) j++;
    38                 array.add(s.substring(i, j));
    39                 i = j;
    40             }
    41         }
    42         return array;
    43     }
    44     public int calculate(String s) {
    45         ArrayList<String> array = split(s);
    46         Stack<Character> opStack = new Stack<Character>();
    47         Stack<Long> numStack = new Stack<Long>();
    48         for(int i = 0; i < array.size(); i++){
    49             if(isNum(array.get(i))){
    50                 numStack.push(Long.parseLong(array.get(i)));
    51             }else{
    52                 char ch = array.get(i).charAt(0);
    53                 if(ch == '('){
    54                     opStack.push(ch);
    55                 }else if(ch == ')'){
    56                     while(opStack.peek() != '('){
    57                         popOpAndTwoNums(opStack, numStack);
    58                     }
    59                     opStack.pop();
    60                 }else{
    61                     if(opStack.size() > 0 && isOp(opStack.peek()) && priority(ch) <= priority(opStack.peek())){
    62                         popOpAndTwoNums(opStack, numStack);
    63                     }
    64                     opStack.push(ch);
    65                 }
    66             }
    67         }
    68         while(numStack.size() > 1){
    69             popOpAndTwoNums(opStack, numStack);
    70         }
    71         return numStack.peek().intValue();
    72     }
    73 }

转载于:https://www.cnblogs.com/liqiniuniu/p/11024229.html