leetcode 27. Remove element

Remove element

Given an array nums and a value val, remove all instances of that value in-place and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn’t matter what you leave beyond the new length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means a modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}

重点：不分配附加的空间给另外一个array，需要通过modifying the input array in-place with O(1) extra memory.

双指针法

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        l,r=0,len(nums)-1
        if len(nums)==0:
            return 0

        while l<r:
        
            while (l<r and nums[l]!=val):
                l+=1
            while (l<r and nums[r]==val):
                r-=1
        
            nums[l],nums[r]=nums[r],nums[l]

        return l if nums[l]==val else l+1

记录
（1）双指针法，设定两个指针，一个从左往右移动，如果遇到第一个val，就停止，另外一个指针从右开始往左移动，遇到第一个不是val的就停止。然后两个元素交换。
（2）移动到两个指针到一起后停止。

我犯的错误
（1）while的条件确定，在里面的while也需要有l<r这项，我把这项弄错了，所以结果一直不对。如果没有这个，也许会导致l>r，就出错了。
（2）细节部分，比如r的初始值是len(nums)-1而不是len(nums)

最后

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