问题描述：

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.

Example 2：

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.

思路：

第一版没有采用java中的封装库，直接写逻辑。使用两个指针，用来指示非val和val，示例如下：

nums=[0，1，2，2，3，0，4]      val = 2      使用index作为非val指示，val_index作为val指示，

第一步：val_index=2, index=4  交换两者位置    nums=[0, 1, 3, 2, 2, 0, 4]

第二步：val_index=3, index=5  交换两者位置    nums=[0, 1, 3, 0, 2, 2, 4]

第三步         val_index=4, index=6  交换两者位置    nums=[0, 1, 3, 0, 4, 2, 2]

当index移动到nums.length-1时说明已经全部交换完成，输出非val数据多少，即val_index的值，代码如下

package leetcode_easy_27_RemoveElement;

class Solution {
    public int removeElement(int[] nums, int val) {
		if(nums.length==0)
			return 0;
        if(nums.length==1)
        	if(nums[0]==val)
        		return 0;
        	else 
        		return 1;
		int val_index = find_val_index(nums, 0, val);
		if(val_index==-1)
			return nums.length;
		int nums_index = find_nums_index(nums, val_index, val);
		if(nums_index==-1)
			return find_val_index(nums, 0, val);
	
		
		while(nums_index!=nums.length)
		{
			
			exchange_nums(nums, val_index, nums_index);
			val_index = find_val_index(nums, val_index, val);
			nums_index = find_nums_index(nums, nums_index, val);
			if(nums_index==-1||val_index==-1)
				break;
		}
		return val_index;
    }
    
    
    
    public void exchange_nums(int[] nums, int fir_idx,int sec_idx)
    {
    	int tmp ;
    	tmp = nums[fir_idx];
    	nums[fir_idx] = nums[sec_idx];
    	nums[sec_idx] = tmp;
    }
    
    /*
     * 找到初始位置
     */
    public int find_val_index(int[] nums, int val_index, int val)
    {
    	for(int i = val_index;i<nums.length;i++)
    	{
    		if(nums[i] == val)
    			{
    			 return i;
    			}
    	}
    	return -1;
    }
    
    public int find_nums_index(int[] nums, int index, int val)
    {
    	for(int i = index;i<nums.length;i++)
    	{
    		if(nums[i] != val)
    			return i;
    	}
    	return -1;
    }   
    
    public static void main(String[] args) {
		Solution solution = new Solution();
		int[] nums =  {4,5};
		System.out.println(solution.removeElement(nums, 5));
		for(int i:nums){
			System.out.print(i);
		}
	}
}

代码是通过了，但是效率非常之低。。于是选择使用java的一些库函数来提高效率

public int removeElement(int[] nums, int val) {
    int i = 0;
    int n = nums.length;
    while (i < n) {
        if (nums[i] == val) {
            nums[i] = nums[n - 1];
            // reduce array size by one
            n--;
        } else {
            i++;
        }
    }
    return n;
}

看了别人的solution之后发现，他们去除完element之后，数字出现的顺序变了/