LeetCode#27 Remove Element

Problem Definition:

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

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

 

Solution:

 

1）一种最简单的思想，花一点空间。

1 def removeElement(nums, val):
2         nu=[]
3         for n in nums:
4             if n!=val:
5                 nu+=n,
6         nums[:]=nu[:]
7         return len(nums)

 

2）两头扫描，替换。

 1 def removeElement(nums, val):
 2         p,q=0,len(nums)-1
 3         while p<=q:
 4             if nums[p]!=val:
 5                 p+=1
 6             elif nums[q]==val:
 7                 q-=1
 8             else:
 9                 nums[p],nums[q]=nums[q],nums[p]
10                 p+=1
11                 q-=1
12         nums[:]=nums[:q+1]
13         return len(nums)

 

3）计数 1。

1 def removeElement(nums, val):
2         m=0    #计应保留的元素数
3         for e in nums:
4             if e!=val:
5                 nums[m]=e
6                 m+=1
7         nums[:]=nums[:m]
8         return m

 

4）计数 2。

 1 def removeElement(nums, val):
 2         m=0    #计应删除的元素数
 3         for i,e in enumerate(nums):
 4             if e==val:
 5                 m+=1
 6             else:
 7                 nums[i-m]=e
 8         ol=len(nums)
 9         nums[:]=nums[:ol-m]
10         return len(nums)