LeetCode-27. Remove Element

题目：

Given an array and a value, 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.

Example:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

solution：

解决方法比较耿直，nums.remove(val)然后返回新的长度即可

class Solution:
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        while val in nums:
            nums.remove(val)

        return len(nums)

在discuss里还有一种解法：

Starting from the left every time we find a value that is the target value we swap it out with an item starting from the right. We decrement end each time as we know that the final item is the target value and only increment start once we know the value is ok. Once start reaches end we know all items after that point are the target value so we can stop there.

从左边开始，每次我们找到一个目标值，我们用从右边开始的项来交换它。然后递减，直至找完。

  def removeElement(self, nums, val):
    start, end = 0, len(nums) - 1
    while start <= end:
        if nums[start] == val:
            nums[start], nums[end], end = nums[end], nums[start], end - 1
        else:
            start +=1
    return start