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| 1 | +package easy; |
| 2 | + |
| 3 | +import utils.CommonUtils; |
| 4 | + |
| 5 | +/**283. Move Zeroes QuestionEditorial Solution My Submissions |
| 6 | +Total Accepted: 105705 |
| 7 | +Total Submissions: 231420 |
| 8 | +Difficulty: Easy |
| 9 | +Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements. |
| 10 | +
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| 11 | +For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0]. |
| 12 | +
|
| 13 | +Note: |
| 14 | +You must do this in-place without making a copy of the array. |
| 15 | +Minimize the total number of operations.*/ |
| 16 | +public class MoveZeroes { |
| 17 | + public void moveZeroes_Editorial_solution2(int[] nums){ |
| 18 | + //this solutoin is the most optimal since it minimizes the number of operations |
| 19 | + //the idea is to swap the non-zero element to the first zero number position |
| 20 | + for(int i = 0, j = 0; i < nums.length && j < nums.length; i++){ |
| 21 | + if(nums[i] != 0){ |
| 22 | + int temp = nums[i]; |
| 23 | + nums[i] = nums[j]; |
| 24 | + nums[j] = temp; |
| 25 | + j++; |
| 26 | + } |
| 27 | + } |
| 28 | + } |
| 29 | + |
| 30 | + public void moveZeroes_Editorial_solution1(int[] nums){ |
| 31 | + //keep the last non-zero index and keep overwriting it, then append zeroes to fill the end |
| 32 | + int j = 0, i = 0; |
| 33 | + for(; j < nums.length; j++){ |
| 34 | + if(nums[j] != 0){ |
| 35 | + nums[i++] = nums[j]; |
| 36 | + } |
| 37 | + } |
| 38 | + for(; i < nums.length; i++){ |
| 39 | + nums[i] = 0; |
| 40 | + } |
| 41 | + } |
| 42 | + |
| 43 | + //then I came up with this solution and got it AC'ed! Cheers! |
| 44 | + //basically, find the next non-zero number and swap it with the current zero number |
| 45 | + //Apparently it's not the most optimal, since this is basically an O(n^2) solution, then I turned to Editorial solutions |
| 46 | + public void moveZeroes(int[] nums){ |
| 47 | + for(int i = 0; i < nums.length-1; i++){ |
| 48 | + if(nums[i] == 0){ |
| 49 | + int j = i+1; |
| 50 | + while(j < nums.length && nums[j] == 0){ |
| 51 | + j++; |
| 52 | + } |
| 53 | + if(j >= nums.length) return; |
| 54 | + else { |
| 55 | + int temp = nums[j]; |
| 56 | + nums[j] = nums[i]; |
| 57 | + nums[i] = temp; |
| 58 | + } |
| 59 | + } |
| 60 | + } |
| 61 | + } |
| 62 | + |
| 63 | + //this approach won't preserve the relative order of the non-zero numbers |
| 64 | + public void moveZeroes_1st_attempt(int[] nums) { |
| 65 | + int i = 0, j = nums.length-1; |
| 66 | + while(i < j){ |
| 67 | + if(nums[i] == 0){ |
| 68 | + int temp = nums[j]; |
| 69 | + nums[j] = nums[i]; |
| 70 | + nums[i] = temp; |
| 71 | + j--; |
| 72 | + } else { |
| 73 | + i++; |
| 74 | + } |
| 75 | + } |
| 76 | + CommonUtils.printArray(nums); |
| 77 | + } |
| 78 | + |
| 79 | + public static void main(String...strings){ |
| 80 | + MoveZeroes test = new MoveZeroes(); |
| 81 | + int[] nums = new int[]{0,1,0,3,12}; |
| 82 | + test.moveZeroes_Editorial_solution2(nums); |
| 83 | + } |
| 84 | +} |
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