[Swift]LeetCode283. 移动零 | Move Zeroes
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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.
Example:
Input:[0,1,0,3,12]
Output:[1,3,12,0,0]
Note:
- You must do this in-place without making a copy of the array.
- Minimize the total number of operations.
给定一个数组 nums
,编写一个函数将所有 0
移动到数组的末尾,同时保持非零元素的相对顺序。
示例:
输入:[0,1,0,3,12]
输出:[1,3,12,0,0]
说明:
- 必须在原数组上操作,不能拷贝额外的数组。
- 尽量减少操作次数。
16ms
1 class Solution { 2 func moveZeroes(_ nums: inout [Int]) { 3 //0要后移,反意是非0元素前移。 4 if nums == nil || nums.count == 0 5 { 6 return 7 } 8 //记录非0元素开始位置 9 var j:Int = 0 10 for i in 0..<nums.count 11 { 12 if nums[i] != 0 13 { 14 nums[j] = nums[i] 15 j += 1 16 } 17 } 18 while(j < nums.count) 19 { 20 nums[j] = 0 21 j += 1 22 } 23 } 24 }
16ms
1 class Solution { 2 func moveZeroes(_ nums: inout [Int]) { 3 var firstPointer = 0 4 var secondPointer = 0 5 6 while secondPointer < nums.count { 7 if nums[firstPointer] == 0 { 8 if nums[secondPointer] != 0 { 9 swap(&nums, firstPointer, secondPointer) 10 } else { 11 secondPointer += 1 12 } 13 } else { 14 secondPointer += 1 15 firstPointer += 1 16 } 17 } 18 } 19 20 func swap(_ nums: inout [Int], _ firstIndex: Int, _ secondIndex: Int) { 21 var temp = nums[firstIndex] 22 nums[firstIndex] = nums[secondIndex] 23 nums[secondIndex] = temp 24 } 25 }
16ms
1 class Solution { 2 func moveZeroes(_ nums: inout [Int]) { 3 var index1 = 0 4 var index2 = nums.count 5 while index1 <= index2 - 1 { 6 if nums[index1] == 0 { 7 for _ in 0..<(index2-index1) { 8 index2 -= 1 9 if nums[index2] != 0 { 10 nums.remove(at: index1) 11 nums.append(0) 12 break; 13 } 14 } 15 }else { 16 index1 += 1 17 } 18 } 19 } 20 }
20ms
1 class Solution { 2 func moveZeroes(_ nums: inout [Int]) { 3 var removedCount = 0 4 for (idx, val) in nums.enumerated() { 5 if val == 0 { 6 nums.remove(at: idx-removedCount) 7 removedCount += 1 8 nums.append(0) 9 } 10 } 11 } 12 }