LeetCode Easy: 27. Remove Element

一、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.

不借助新的空间，删除给定数组中的与给定的 val 相同的元素，返回新数组的长度，新数组的末尾是什么不关心。

二、思路

首先不考虑鲁棒性，定义两个指针，其中一个指针遍历整个数组，另外一个指针负责计数即当前数组元素不等于给定 val 时做加 1操作 ，同时将该元素前移。大的考虑就这些。

三、代码

#coding:utf-8
def removeElement(nums, val):
    """
    :type nums: List[int]
    :type val: int
    :rtype: int
    """
    numslenth = len(nums)
    j=0
    # if numslenth == 0:
    #     return 0
    # if numslenth == 1 and val == nums[0] :
    #     return 0
    # if numslenth == 1 and val != nums[0] :
    #     return 1
    for i in range(0,numslenth):
        if nums[i] != val:
            nums[j] = nums[i]
            j += 1
    print(j,"\n",nums)
    return j
if __name__ == '__main__':
    nums = [5]
    val = 3
    removeElement(nums,val)