thanhnguyennguyen /
cff
| Total Complexity | 8 |
| Complexity/F | 8 |
| Lines of Code | 42 |
| Function Count | 1 |
| Duplicated Lines | 23 |
| Ratio | 54.76 % |
| Changes | 0 | ||
Duplicate code is one of the most pungent code smells. A rule that is often used is to re-structure code once it is duplicated in three or more places.
Common duplication problems, and corresponding solutions are:
| 1 | const Stack = require('../index.js'); |
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| 2 | const operators = { |
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| 3 | "+": 1, |
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| 4 | "-": 1, |
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| 5 | "*": 2, |
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| 6 | "/": 2, |
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| 7 | "(": 3, |
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| 8 | ")": 3, |
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| 9 | } |
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| 10 | |||
| 11 | const infixToPrefix = (formula) => { |
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| 12 | let result = "" |
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| 13 | let stack = new Stack() |
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| 14 | View Code Duplication | for (let i = formula.length - 1; i >= 0; i--) { |
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| 15 | let c = formula[i] |
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| 16 | if (!operators[c]) { |
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| 17 | result += c |
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| 18 | continue |
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| 19 | } |
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| 20 | if (c === '(') { |
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| 21 | while (!stack.isEmpty() && stack.top() !== ')') { |
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| 22 | result += stack.pop() |
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| 23 | } |
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| 24 | stack.pop() // pop ')' |
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| 25 | continue |
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| 26 | } |
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| 27 | if (stack.isEmpty() || operators[c] >= operators[stack.top()] || stack.top() === ')') { |
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| 28 | stack.push(c) |
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| 29 | continue |
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| 30 | } |
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| 31 | while (!stack.isEmpty() && operators[stack.top()] > operators[c]) { |
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| 32 | let p = stack.pop() |
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| 33 | result += p |
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| 34 | } |
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| 35 | stack.push(c) |
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| 36 | } |
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| 37 | while (!stack.isEmpty()) { |
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| 38 | result += stack.pop() |
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| 39 | } |
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| 40 | return result.split('').reverse().join('') |
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| 41 | } |
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| 42 | module.exports = infixToPrefix |
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| 43 |
