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.
Example 1:
Given nums = [3,2,2,3], val = 3,
Example 2:
Given nums = [0,1,2,2,3,0,4,2], val = 2,
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 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)
翻译：
给定一个数组和一个定值，删除数组中的所有此值并返回一个新的数组长度
不要申明额外的数组空间，必须保证算法复杂度为O(1)
数组的顺序无所谓
思路：
这一题和【LeetCode第26题】差不多思路，只不过一个是删除重复值，一个是删除给定值
既然不能申明额外的数组，那只能在原来的数组上做变动
变动前：[1,1,2,3,3]，给定值为：1
变动后：[2,3,3,3,3]
前3个值[2,3,3]就是我们所需要的
代码：
1class Solution { 2 3因为最后一次进入if(nums[i] != val)判断后，还执行了一次j++，所以j的值就已经等于数组长度了，return的时候不需要再+1了
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