数组移动问题

#include<stdio.h>void shiftleft(int *pInOut, int n){int i;int tmp;tmp = *pInOut;for (i = 0; i < n - 1; ++i){*pInOut = *(pInOut + 1);pInOut++;}*pInOut = tmp;}void shiftright(int *pInOut, int n){int i;int tmp;pInOut += n - 1;tmp = *pInOut;for (i = n - 1 ; i > 0; --i){*pInOut = *(pInOut - 1);--pInOut;}*pInOut = tmp;}void shiftN(int *pInOut, int n, int shiftN){int i;if(shiftN > 0){for(i = 0; i < shiftN % n; ++i)shiftleft(pInOut, n);}else{shiftN = -shiftN;for(i = 0; i < shiftN % n; ++i)shiftright(pInOut, n);}}int main(){int array[4] = {1, 2, 3, 4};int array1[3] = {1, 2, 3};int i;shiftN(array, 4, 5);for(i = 0; i < 4; ++i)printf("%d ",array[i]);printf("\n"); shiftN(array1, 3, -1);for(i = 0; i < 3; ++i)printf("%d ",array1[i]);printf("\n");return 0;}

数组移动操作，写一个函数void  shitfN(int *pInOut, int n, int shiftN)，将pInOut指向的数组移动shiftN位，如果shiftN为正数，就左移，如果为负数就右移，n为数组的长度。例如，array[]={1, 2, 3, 4}, shiftN = 1,移动后的结果为，array[] ={ 2, 3, 4, 1}  

如果array[]={1, 2, 3},shiftN = -1,移动后的结果为，array[]={3, 1, 2}

时间复杂度为o（ShiftN*n），换个方法思考：

循环左移ShiftN个，效果等价于，前ShiftN个元素逆置，后n-ShiftN个元素逆置，最后n个元素整体逆置。时间复杂度为o(n).