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December 1, 2023 02:22
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[Destructuring Assignment] #JavaScript #Array #
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| /** | |
| * Function: moveZeros | |
| * Description: This function moves all zeros in an array to the end of the array, while | |
| * preserving the order of the other elements. Without creating a new Array. | |
| * Note: this uses a common two-pointer method, is O(n) and does not create a new array | |
| * @param {Array} nums - The input array. | |
| * @return {Array} The modified input array. | |
| */ | |
| const moveZeros = (nums) => { | |
| let left = 0 | |
| let right = 0 | |
| while (right < nums.length) { | |
| // If the current element is not zero, | |
| // swap it with the element at the 'left' index | |
| if (nums[right] !== 0) { | |
| // destructuring assignment with swap. | |
| // Doing this way avoids allocation three variables | |
| [nums[left], nums[right]] = [nums[right],nums[left]] | |
| left++; | |
| } | |
| right++; | |
| } | |
| return nums | |
| } | |
| console.log( moveZeros([0,1,2,3]) ) | |
| console.log( moveZeros([1,2,0,3,4]) ) | |
| console.log( moveZeros([1,2,3,4,0]) ) | |
| console.log( moveZeros([1,0,0,4]) ) | |
| console.log( moveZeros([0,2,3,0]) ) | |
| console.log( moveZeros([0]) ) | |
| console.log( moveZeros([1]) ) | |
| console.log( moveZeros([]) ) | |
| /** | |
| * [ 1, 2, 3, 0 ] | |
| * [ 1, 2, 3, 4, 0 ] | |
| * [ 1, 2, 3, 4, 0 ] | |
| * [ 1, 4, 0, 0 ] | |
| * [ 2, 3, 0, 0 ] | |
| * [ 0 ] | |
| * [ 1 ] | |
| * [] | |
| */ |
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