Remove Element

题目要求：

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,

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

It doesn't matter what you leave beyond the returned length.

Example2：

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

It doesn't matter what values are set beyond the returned length.

分析：题目的初衷是让你按照要求修改原数组，最后返回删除指定元素后数组的长度。（这里的删除并不是真的删除了数组中的某个元素，只是用后面的数把前面的数给覆盖罢了，并且题目在“Clarification”中还指出不需理会数组中length后的元素。）那么，我们这里的思路是：如果数组中的某个元素与指定val相等，我们直接让后面的元素顶替上来即可，最后得到的结果是，数组length之前的元素都不包含val。

class Solution {
   public int removeElement(int[] nums, int val) {
    if (nums.length == 0) {
           return 0;
       }
       int len = 0;
       for (int i = 0; i < nums.length; i++) {
           if(nums[i]==val){
               continue;
           }
           nums[len]=nums[i];
           len++;
       }
       return len;
   }
}