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Develop #2448
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| @@ -0,0 +1,23 @@ | ||
| name: Test | ||
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| on: | ||
| pull_request: | ||
| branches: [ master ] | ||
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| jobs: | ||
| build: | ||
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| runs-on: ubuntu-latest | ||
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| strategy: | ||
| matrix: | ||
| node-version: [20.x] | ||
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| steps: | ||
| - uses: actions/checkout@v2 | ||
| - name: Use Node.js ${{ matrix.node-version }} | ||
| uses: actions/setup-node@v1 | ||
| with: | ||
| node-version: ${{ matrix.node-version }} | ||
| - run: npm install | ||
| - run: npm test |
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@@ -4,8 +4,21 @@ | |
| * Implement method Sort | ||
| */ | ||
| function applyCustomSort() { | ||
| [].__proto__.sort2 = function(compareFunction) { | ||
| // write code here | ||
| [].__proto__.sort2 = function ( | ||
| compareFunction = (a, b) => String(a) > String(b), | ||
| ) { | ||
| for (let i = 0; i < this.length - 1; i++) { | ||
| for (let j = 0; j < this.length - 1; j++) { | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Your bubble sort implementation is correct. For a performance improvement, you can optimize the inner loop. After each pass of the outer loop, the largest unsorted element is moved to its correct position. Therefore, the inner loop doesn't need to iterate over the elements that are already sorted at the end of the array. You can adjust the inner loop's condition to |
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| if (compareFunction(this[j], this[j + 1]) > 0) { | ||
| const temp = this[j]; | ||
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| this[j] = this[j + 1]; | ||
| this[j + 1] = temp; | ||
| } | ||
| } | ||
| } | ||
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| return this; | ||
| }; | ||
| } | ||
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The standard
compareFunctionforArray.prototype.sortis expected to return a number (negative, zero, or positive). Your function currently returns a boolean. While this works in your current implementation due to JavaScript's type coercion (truebecomes 1,falsebecomes 0), it's better practice to adhere to the standard. This makes your code more predictable and reusable. Consider returning a number, for example by usingString(a).localeCompare(String(b)).