【LeetCode】Remove Element

版权声明： https://blog.csdn.net/ysq96/article/details/89949875

Given an array nums and a value val, remove all instances of that value in-placeand 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.

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]);
}

题意：这题和之前的有序数组删除重复数字有点类似。这题要求在无序数组中删除掉给定的值，将剩下的数字组成全新的数组，并返回全新数组的大小。另题目要求O（1）的空间复杂度

C++：

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

Python3：

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        j=0
        nums_len=len(nums)
        for i in range(nums_len):
            if nums[i]!=val:
                nums[j]=nums[i]
                j+=1
        return j