PRoblem:
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 in place with constant memory. The order of elements can be changed. It doesn’t matter what you leave beyond the new length. Example: Given input array nums = [3,2,2,3], val = 3 Your function should return length = 2, with the first two elements of nums being 2.
Idea: 方法与26题相似。 另外也可以对原数组进行覆盖赋值，因为判断正误时只会看数组长度new length 前面的那段。
Solution:
class Solution(object): def removeElement(self, nums, val): i=0 lennums = len(nums) while i < lennums: if val == nums[i]: nums.remove(val) lennums -= 1 else: i += 1 return lennums新闻热点
疑难解答
