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].
Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
此题在《剑指Offer》上面遇到过,而且解题思路也还记得,如下:
将数组中前面(n-k)个元素逆序,以及将数组后面的k个元素 逆序,最后将整体逆序即可。
/*
思路:将数组中前面(n-k)个元素逆序,以及将数组后面的k个元素 逆序
最后将整体逆序即可。
*/
void swap(int *a,int *b){
int temp=*a;
*a=*b;
*b=temp;
}
//函数功能:将数组nums进行逆序
void reverse(int *nums,int numsSize){
if(nums==NULL||numsSize<1){
return ;
}
int begin=0;
int end=numsSize-1;
while(begin<end){
swap(&nums[begin],&nums[end]);
begin++;
end--;
}
}
void rotate(int* nums, int numsSize, int k) {
if(nums==NULL||numsSize<1||k<0){
return;
}
k=k%numsSize;
reverse(nums,numsSize-k);
reverse(nums+numsSize-k,k);
reverse(nums,numsSize);
}