删除数组元素和移位，其余的 [英] Remove an array element and shift the remaining ones

问题描述

我如何删除数组的元素和移位剩余的元素了。所以，如果我有一个数组，

How do I remove an element of an array and shift the remaining elements down. So, if I have an array,

array[]={1,2,3,4,5} 

和要删除3等我倒腾休息，

and want to delete 3 and shift the rest so I have,

array[]={1,2,4,5}

我怎么会去这code量最少的？

How would I go about this in the least amount of code?

推荐答案

您只需要重写你的数组中的下一个值删除的内容，传播这一变化，然后记住其中新到底是：

You just need to overwrite what you're deleting with the next value in the array, propagate that change, and then keep in mind where the new end is:

int array[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};

// delete 3 (index 2)
for (int i = 2; i < 8; ++i)
    array[i] = array[i + 1]; // copy next element left

现在您的数组是 {1，2，4，5，6，7，8，9，9} 。您不能删除多余的 9 ，因为这是一个静态大小的数组，你只需要忽略它。这是可以做到的的std ::复制：

Now your array is {1, 2, 4, 5, 6, 7, 8, 9, 9}. You cannot delete the extra 9 since this is a statically-sized array, you just have to ignore it. This can be done with std::copy:

std::copy(array + 3, // copy everything starting here
          array + 9, // and ending here, not including it,
          array + 2) // to this destination

在C ++ 11，使用可以使用的std ::移动（算法超载，不实用过载）来代替。

In C++11, use can use std::move (the algorithm overload, not the utility overload) instead.

更一般地，使用的std ::删除删除匹配值的元素：

More generally, use std::remove to remove elements matching a value:

// remove *all* 3's, return new ending (remaining elements unspecified)
auto arrayEnd = std::remove(std::begin(array), std::end(array), 3);

更普遍，有的std ::的remove_if 。

请注意，使用的std ::矢量＆lt的; INT＆GT; 可能更适合在这里，因为它是一个真正的动态分配调整阵列。 （在这个意义上，要求其尺寸（）反映移除元素。）

Note that the use of std::vector<int> may be more appropriate here, as its a "true" dynamically-allocated resizing array. (In the sense that asking for its size() reflects removed elements.)

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