#27 Remove Element

Description

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.

Examples

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.

解题思路

完全承接26hhhhhhh
直接放代码好了

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

Two Pointers - when elements to remove are rare

照之前又去摸了一下solution
看到一个当元素移动次数特别少的情况下可以用的算法（对这题比较适用）
依次判断当前pointer $i$ 的值是否等于 $val$
如果不等于， $i$++，索引向前移动一位继续判断下一个
如果等于，则把最后一个值赋值给这个位置， $i$ 不动，数组长度 - 1

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;
}