27 Remove Element 移除元素

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.

Example:

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.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// **nums** is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);

// any modification to **nums** in your function would be known by the caller.
// using the length returned by your function, it prints the first **len** elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

题目描述:
给定一个数组 nums和一个值 val，你需要原地移除所有数值等于 val的元素，返回移除后数组的新长度。

不要使用额外的数组空间，你必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成。

元素的顺序可以改变。你不需要考虑数组中超出新长度后面的元素。

示例:

示例 1:

给定 nums = [3,2,2,3], val = 3,

函数应该返回新的长度 2, 并且 nums 中的前两个元素均为 2。

你不需要考虑数组中超出新长度后面的元素。

示例 2:

给定 nums = [0,1,2,2,3,0,4,2], val = 2,

函数应该返回新的长度 5, 并且 nums 中的前五个元素为 0, 1, 3, 0, 4。

注意这五个元素可为任意顺序。

你不需要考虑数组中超出新长度后面的元素。

说明:

为什么返回数值是整数，但输出的答案是数组呢?

请注意，输入数组是以“引用”方式传递的，这意味着在函数里修改输入数组对于调用者是可见的。

你可以想象内部操作如下:

// **nums** 是以“引用”方式传递的。也就是说，不对实参作任何拷贝
int len = removeElement(nums, val);

// 在函数里修改输入数组对于调用者是可见的。
// 根据你的函数返回的长度, 它会打印出数组中**该长度范围内**的所有元素。
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

思路:

当数组为空的时候返回0; 维护一个长度, 遍历数组找到相同值替换.
时间复杂度O(n), 空间复杂度O(1)

代码:
C++:

class Solution 
{
public:
    int removeElement(vector<int>& nums, int val) 
    {
        if (!nums.size()) return 0;
        int result = 0;
        for (int i = 0; i < nums.size(); i++) if (nums[i] != val) nums[result++] = nums[i];
        return result;
    }
};

Java:

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

Python:

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        for num in nums[:]:
            if num == val:
                nums.remove(num)
        return len(nums)