27. Remove elements

一、题意
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.

2. Analysis and Answers

 public int removeElement(int[] nums, int val) {
        if(nums.length == 0)
            return 0;
        int p = 0,q = 0;
        int num = 0;
        while(p<nums.length && q<nums.length){
            while(p < nums.length && nums[p] != val){         
                p++;
            }
            q = p;
            while(q < nums.length && nums[q] == val){
                q++;
            }

            if(p < nums.length && q < nums.length){
                nums[p] = nums[p] ^ nums[q];
                nums[q] = nums[p] ^ nums[q];
                nums[p] = nums[p] ^ nums[q];    
            }
            p++;
        }

        for(int i=nums.length-1;i>=0 && nums[i]==val;i--){
            num++;
        }
        return nums.length - num;
    }

Note:
(1) while(p < nums.length && nums[p] != val){ p++; }
, must have p < nums.lengththis judgment. Because there is p++ inside, the last time the number of p is nums.length, and it can also enter the while judgment at this time, the array is out of bounds!
(2) In-place transformation, using the XOR operation!
(3) Use fast and slow pointers, the slow pointer points to a value equal to val, and the fast pointer is equal to the value behind this value that is not equal to val!

Others' method:

    public int removeElement(int[] nums, int val) {
        if(nums.length == 0)
            return 0;
        int count = 0;
        for(int i=0;i<nums.length;i++){
            if(nums[i] != val)
                nums[count++] = nums[i];
        }
        return count;
    }

Although others look simple, it seems that mine is faster than it! Not necessarily, maybe the server pressure was low at that time.