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165 lines
3.6 KiB
Markdown
Executable File
165 lines
3.6 KiB
Markdown
Executable File
# [224. Basic Calculator](https://leetcode.com/problems/basic-calculator/)
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## 题目
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Implement a basic calculator to evaluate a simple expression string.
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The expression string may contain open `(` and closing parentheses `)`, the plus `+` or minus sign `-`, **non-negative** integers and empty spaces .
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**Example 1**:
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Input: "1 + 1"
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Output: 2
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**Example 2**:
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Input: " 2-1 + 2 "
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Output: 3
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**Example 3**:
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Input: "(1+(4+5+2)-3)+(6+8)"
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Output: 23
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**Note**:
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- You may assume that the given expression is always valid.
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- **Do not** use the `eval` built-in library function.
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## 题目大意
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实现一个基本的计算器来计算一个简单的字符串表达式的值。字符串表达式可以包含左括号 ( ，右括号 )，加号 + ，减号 -，非负整数和空格 。
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## 解题思路
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- 注意点一：算式中有空格，需要跳过
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- 注意点二：算式中会出现负数，负负得正的情况需要特殊处理，所以需要记录每次计算出来的符号
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## 代码
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```go
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package leetcode
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import (
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"container/list"
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"fmt"
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"strconv"
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)
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// 解法一
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func calculate(s string) int {
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i, stack, result, sign := 0, list.New(), 0, 1 // 记录加减状态
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for i < len(s) {
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if s[i] == ' ' {
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i++
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} else if s[i] <= '9' && s[i] >= '0' { // 获取一段数字
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base, v := 10, int(s[i]-'0')
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for i+1 < len(s) && s[i+1] <= '9' && s[i+1] >= '0' {
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v = v*base + int(s[i+1]-'0')
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i++
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}
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result += v * sign
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i++
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} else if s[i] == '+' {
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sign = 1
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i++
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} else if s[i] == '-' {
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sign = -1
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i++
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} else if s[i] == '(' { // 把之前计算结果及加减状态压栈，开始新的计算
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stack.PushBack(result)
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stack.PushBack(sign)
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result = 0
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sign = 1
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i++
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} else if s[i] == ')' { // 新的计算结果 * 前一个加减状态 + 之前计算结果
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result = result*stack.Remove(stack.Back()).(int) + stack.Remove(stack.Back()).(int)
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i++
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}
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}
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return result
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}
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// 解法二
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func calculate1(s string) int {
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stack := []byte{}
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for i := 0; i < len(s); i++ {
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if s[i] == ' ' {
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continue
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} else if s[i] == ')' {
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tmp, index := "", len(stack)-1
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for ; index >= 0; index-- {
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if stack[index] == '(' {
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break
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}
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}
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tmp = string(stack[index+1:])
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stack = stack[:index]
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res := strconv.Itoa(calculateStr(tmp))
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for j := 0; j < len(res); j++ {
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stack = append(stack, res[j])
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}
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} else {
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stack = append(stack, s[i])
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}
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}
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fmt.Printf("stack = %v\n", string(stack))
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return calculateStr(string(stack))
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}
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func calculateStr(str string) int {
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s, nums, tmpStr, res := []byte{}, []int{}, "", 0
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// 处理符号的问题，++得+，--得+，+-、-+得-
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for i := 0; i < len(str); i++ {
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if len(s) > 0 && s[len(s)-1] == '+' && str[i] == '+' {
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continue
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} else if len(s) > 0 && s[len(s)-1] == '+' && str[i] == '-' {
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s[len(s)-1] = '-'
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} else if len(s) > 0 && s[len(s)-1] == '-' && str[i] == '+' {
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continue
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} else if len(s) > 0 && s[len(s)-1] == '-' && str[i] == '-' {
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s[len(s)-1] = '+'
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} else {
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s = append(s, str[i])
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}
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}
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str = string(s)
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s = []byte{}
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for i := 0; i < len(str); i++ {
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if isDigital(str[i]) {
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tmpStr += string(str[i])
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} else {
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num, _ := strconv.Atoi(tmpStr)
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nums = append(nums, num)
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tmpStr = ""
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s = append(s, str[i])
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}
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}
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if tmpStr != "" {
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num, _ := strconv.Atoi(tmpStr)
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nums = append(nums, num)
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}
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res = nums[0]
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for i := 0; i < len(s); i++ {
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if s[i] == '+' {
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res += nums[i+1]
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} else {
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res -= nums[i+1]
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}
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}
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fmt.Printf("s = %v nums = %v res = %v\n", string(s), nums, res)
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return res
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}
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func isDigital(v byte) bool {
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if v >= '0' && v <= '9' {
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return true
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}
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return false
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}
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``` |