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

双指针法 用一根指针P1 代表满足不等于VAL的值的part，用P2代表满足等于VAL的part，注意P1与P2都不是包括在这两个部分的，换句话说满足不等于VAL的part在0-(P1-1),满足等于VAL的值在P2+1---end.所以边界应该是P1==P2。此时左侧是不等于VAL的，右侧是等于VAL的，只剩下P1==P2处没有检验。

class Solution {
    public int removeElement(int[] nums, int val) {
        if(nums.length==1)
        {
            return nums[0]==val? 0:1; 
        }
        int pos1 =0;
        int pos2 = nums.length-1;
        while(pos1<=pos2)
        {
            while(pos1<=pos2&&nums[pos1]!=val)
                pos1++;
            //find one
            if(pos1==pos2+1)
                return pos2+1;
            else
            {
                int temp = nums[pos1];
                nums[pos1]=nums[pos2];
                nums[pos2]=temp;
                pos2--;
            }
        }
        return pos1;
    }
}