C++中怎么利用LeetCode移除元素

這篇文章給大家介紹C++中怎么利用LeetCode移除元素，內容非常詳細，感興趣的小伙伴們可以參考借鑒，希望對大家能有所幫助。

[LeetCode] 27. Remove Element 移除元素

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]);
}

這道題讓我們移除一個數組中和給定值相同的數字，并返回新的數組的長度。是一道比較容易的題，只需要一個變量用來計數，然后遍歷原數組，如果當前的值和給定值不同，就把當前值覆蓋計數變量的位置，并將計數變量加1。代碼如下：

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

關于C++中怎么利用LeetCode移除元素就分享到這里了，希望以上內容可以對大家有一定的幫助，可以學到更多知識。如果覺得文章不錯，可以把它分享出去讓更多的人看到。