HappyLeetcode50:Rotate Array

Rotate an array of n elements to the right by k steps.

For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

这道题虽然做了出来，但是显然不是最优解。显然空间要求没有达到。

有一点要尤为注意，在一开始的时候我就默认k是小于n的，实际上k可以大于n，所以在修改代码之后，我们所使用的k要对n取余。

我的思路是，把数组中前面的值先挪到后面去。后面的值在被替换之前先存进一个vector里面去，待前面的值安排完毕之后，把vector之中的数据取出，放在新数组的前面。

代码如下:

class Solution {
public:void rotate(int nums[], int n, int k) {if (nums == NULL || n <= 0 )return;k = k%n;vector<int> TempSave;TempSave.reserve(k);int count = 0;//记录已经推进了多少个数据for (int i = n - k - 1; i >= 0; --i){if (count < k){TempSave.push_back(nums[i + k]);count++;}nums[i + k] = nums[i];}int Temp = TempSave.size();if (TempSave.size() < k){for (int i = k - 1; i >= Temp; --i)TempSave.push_back(nums[i]);}for (int i = 0; i < k; ++i){nums[i] = TempSave.back();TempSave.pop_back();}    }
};

其实后来想想还是有更简单一些的方法，比如先把数组复制一遍，再根据情况裁剪出你需要的那一部分数组出来。

class Solution {
public:void rotate(int nums[], int n, int k) {k = k % n;int* NewNums =new int[n * 2];for (int i = 0; i < n ; ++i){NewNums[i] = nums[i];NewNums[i + n] = nums[i];}//int res = n - k;for (int i = 0; i < n; ++i){nums[i] = NewNums[i + n - k];}}
};

代码量明显短了不少。

转载于:https://www.cnblogs.com/chengxuyuanxiaowang/p/4302957.html

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