LeetCode刷题——Array篇(Move Zeroes和Remove Element)

最新推荐文章于 2025-02-04 22:28:17 发布

原创最新推荐文章于 2025-02-04 22:28:17 发布 · 276 阅读

0 ·

CC 4.0 BY-SA版权

Array 专栏收录该内容

16 篇文章

订阅专栏

本文介绍了解决两个典型数组操作问题的方法：一是将数组中所有0移动到末尾的同时保持非零元素的相对顺序；二是如何在原地移除指定值的所有实例，并返回新的长度。提供了详细的算法思路及C++实现代码。

Move Zeroes

Given an array nums, write a function to move all 0’s to the end of it while maintaining the relative order of the non-zero elements.

For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].

Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.

Remove Element

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.
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]);
}

思路：
1、一开始想用一个指针来解决这个问题，发现有点困难，后来采用两个指针的方式，一个指针指向每个元素，另一个指向移动后的新元素。

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int count = 0;
        int j = 0;
        for (int i = 0; i < nums.size(); i++)
        {
            if (nums[i ] == val  )
            {
                count++;
            }
            else
            {
                nums[j] = nums[i];
                j++;
            }
        }
        cout << endl << count;
        nums.erase(nums.end()-count, nums.end());

        for(int i = 0; i < nums.size() ; i++)
            cout << nums[i] << "-";
        return  nums.size();
    }
};