Go implements leetcode rotation array

The goal is ： Rotated array

Give you an array , Rotate the elements in the array to the right k A place , among k It is a non negative number .

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Result output ：

Let's start with the simplest loop change , Execution result display timeout , Optimized by the bank ,k After taking the mold, it still times out , Consider that it should be a waste of time to operate the array .

func rotate(nums []int, k int)[]int  {
    
	for i :=0;i<k;i++{
    
		a := nums[(len(nums)-1)]
		for j := len(nums)-1;j>0;j--{
    
			nums[j] = nums[j-1]
		}
		nums[0]=a
	}
	return  nums
}

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Use temporary arrays , Calculate the position of each element , Then add it back to the new array .

There's a problem here ,leetcode The detection fails all the time , There should be no problem checking the logic , Then run the display on your computer without any problem . This is because he directly tested nums Array , Instead of line checking the return value .

func rotate(nums []int, k int)[]int  {
    
	var nums1 = make([]int,len(nums))
	k = k%len(nums)
	for j :=0;j<len(nums);j++ {
    
		if j+k<len(nums) {
    
			nums1[j+k]=nums[j]
		}else {
    
			nums1[(j+k)%len(nums)]=nums[j]
		}
	}
	return  nums1
}

After modification

Execution time ：28 ms, In all Go Defeated in submission 26.28% Users of

Memory consumption ：7.9 MB, In all Go Defeated in submission 45.76% Users of

func rotate(nums []int, k int){
    
	var nums1 = make([]int,len(nums))
	k %= len(nums)
	for j :=0;j<len(nums);j++ {
    
		if j+k<len(nums) {
    
			nums1[j+k]=nums[j]
		}else {
    
			nums1[(j+k)%len(nums)]=nums[j]
		}
	}
    for j :=0;j<len(nums);j++ {
    
        nums[j] =nums1[j]
    }
}

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