这篇文章给大家介绍C++中怎么利用LeetCode移除元素，内容非常详细，感兴趣的小伙伴们可以参考借鉴，希望对大家能有所帮助。
[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]);
}
这道题让我们移除一个数组中和给定值相同的数字，并返回新的数组的长度。是一道比较容易的题，只需要一个变量用来计数，然后遍历原数组，如果当前的值和给定值不同，就把当前值覆盖计数变量的位置，并将计数变量加1。代码如下：
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; }};关于C++中怎么利用LeetCode移除元素就分享到这里了，希望以上内容可以对大家有一定的帮助，可以学到更多知识。如果觉得文章不错，可以把它分享出去让更多的人看到。
