Leetcode: Basic Calculator

最新推荐文章于 2021-02-18 15:49:54 发布

转载最新推荐文章于 2021-02-18 15:49:54 发布 · 68 阅读

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原文链接：http://www.cnblogs.com/EdwardLiu/p/5058595.html

本文介绍了一种使用栈来解析和计算包含括号、加减运算及空格的简单数学表达式的方法。通过两次遍历字符串的方式，实现了有效的计算逻辑。

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 .

You may assume that the given expression is always valid.

Some examples:
"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23

https://discuss.leetcode.com/topic/15816/iterative-java-solution-with-stack

digit: it should be one digit from the current number
'+': number is over, we can add the previous number and start a new number
'-': same as above
'(': push the previous result and the sign into the stack, set result to 0, just calculate the new result within the parenthesis.
')': pop out the top two numbers from stack, first one is the sign before this pair of parenthesis, second is the temporary result before this pair of parenthesis. We add them together.

第二遍做法：遇到digit直接走到末尾，并且直接利用sign进行计算，这样比较方便

 1 public static int calculate(String s) {
 2     int len = s.length(), sign = 1, result = 0;
 3     Stack<Integer> stack = new Stack<Integer>();
 4     for (int i = 0; i < len; i++) {
 5         if (Character.isDigit(s.charAt(i))) {
 6             int sum = s.charAt(i) - '0';
 7             while (i + 1 < len && Character.isDigit(s.charAt(i + 1))) {
 8                 sum = sum * 10 + s.charAt(i + 1) - '0';
 9                 i++;
10             }
11             result += sum * sign;
12         } else if (s.charAt(i) == '+')
13             sign = 1;
14         else if (s.charAt(i) == '-')
15             sign = -1;
16         else if (s.charAt(i) == '(') {
17             stack.push(result);
18             stack.push(sign);
19             result = 0;
20             sign = 1;
21         } else if (s.charAt(i) == ')') {
22             result = result * stack.pop() + stack.pop(); // notice 这里不是+=
23         }
24 
25     }
26     return result;
27 }

第一遍做法：

 1 public class Solution {
 2     public int calculate(String s) {
 3         if (s==null || s.length()==0) return 0;
 4         Stack<Integer> st = new Stack<Integer>();
 5         int sum = 0;
 6         int num = 0;
 7         int sign = 1;
 8         for (int i=0; i<s.length(); i++) {
 9             char cur = s.charAt(i);
10             if (Character.isDigit(cur)) {
11                 num = num*10 + (int)(cur-'0');
12             }
13             else if (cur == '+') {
14                 sum += sign*num;
15                 num = 0;
16                 sign = 1;
17             }
18             else if (cur == '-') {
19                 sum += sign*num;
20                 num = 0;
21                 sign = -1;
22             }
23             else if (cur == '(') {
24                 st.push(sum);
25                 st.push(sign);
26                 sum = 0;
27                 sign = 1;
28             }
29             else if (cur == ')') {
30                 sum += sign*num;
31                 num = 0;
32                 sum *= st.pop();
33                 sum += st.pop();
34             }
35 
36         }
37         if (num != 0) sum += sign*num;
38         return sum;
39     }
40 }

转载于:https://www.cnblogs.com/EdwardLiu/p/5058595.html