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80 lines
2.2 KiB
Markdown
80 lines
2.2 KiB
Markdown
# [227. Basic Calculator II](https://leetcode.com/problems/basic-calculator-ii/)
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## 题目
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Given a string `s` which represents an expression, *evaluate this expression and return its value*.
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The integer division should truncate toward zero.
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**Example 1:**
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```
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Input: s = "3+2*2"
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Output: 7
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```
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**Example 2:**
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```
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Input: s = " 3/2 "
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Output: 1
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```
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**Example 3:**
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```
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Input: s = " 3+5 / 2 "
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Output: 5
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```
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**Constraints:**
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- `1 <= s.length <= 3 * 10^5`
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- `s` consists of integers and operators `('+', '-', '*', '/')` separated by some number of spaces.
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- `s` represents **a valid expression**.
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- All the integers in the expression are non-negative integers in the range `[0, 2^31 - 1]`.
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- The answer is **guaranteed** to fit in a **32-bit integer**.
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## 题目大意
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给你一个字符串表达式 `s` ，请你实现一个基本计算器来计算并返回它的值。整数除法仅保留整数部分。
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## 解题思路
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- 这道题是第 224 题的加强版。第 224 题中只有加减运算和括号，这一题增加了乘除运算。由于乘除运算的优先级高于加减。所以先计算乘除运算，将算出来的结果再替换回原来的算式中。最后只剩下加减运算，于是题目降级成了第 224 题。
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- 把加减运算符号后面的数字压入栈中，遇到乘除运算，直接将它与栈顶的元素计算，并将计算后的结果放入栈顶。若读到一个运算符，或者遍历到字符串末尾，即认为是遍历到了数字末尾。处理完该数字后，更新 `preSign` 为当前遍历的字符。遍历完字符串 `s` 后，将栈中元素累加，即为该字符串表达式的值。时间复杂度 O(n)，空间复杂度 O(n)。
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## 代码
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```go
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package leetcode
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func calculate(s string) int {
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stack, preSign, num, res := []int{}, '+', 0, 0
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for i, ch := range s {
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isDigit := '0' <= ch && ch <= '9'
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if isDigit {
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num = num*10 + int(ch-'0')
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}
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if !isDigit && ch != ' ' || i == len(s)-1 {
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switch preSign {
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case '+':
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stack = append(stack, num)
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case '-':
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stack = append(stack, -num)
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case '*':
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stack[len(stack)-1] *= num
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default:
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stack[len(stack)-1] /= num
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}
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preSign = ch
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num = 0
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}
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}
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for _, v := range stack {
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res += v
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}
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return res
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}
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``` |