给定一个数组，将数组中的元素向右移动 k 个位置。
示例 1:
输入: [1,2,3,4,5,6,7] 和 k = 3
输出: [5,6,7,1,2,3,4]
解释:
向右旋转 1 步: [7,1,2,3,4,5,6]
向右旋转 2 步: [6,7,1,2,3,4,5]
向右旋转 3 步: [5,6,7,1,2,3,4]
示例 2:
输入: [-1,-100,3,99] 和 k = 2
输出: [3,99,-1,-100]
解释:
向右旋转 1 步: [99,-1,-100,3]
向右旋转 2 步: [3,99,-1,-100]三次翻转法
- 将数组第i ( i∈[0,n-1-k] ) 项进行对称翻转
- 将数组第i ( i∈[n-k,n-1] ) 项进行对称翻转
- 将数组第i ( i∈[0,n-1] ) 项进行对称翻转
在lua源码中lua_rotate就是用这个方式实现旋转的：
/*
** Reverse the stack segment from 'from' to 'to'
** (auxiliary to 'lua_rotate')
** Note that we move(copy) only the value inside the stack.
** (We do not move additional fields that may exist.)
*/
l_sinline void reverse (lua_State *L, StkId from, StkId to) {
for (; from < to; from++, to--) {
TValue temp;
setobj(L, &temp, s2v(from));
setobjs2s(L, from, to);
setobj2s(L, to, &temp);
}
}
/*
** Let x = AB, where A is a prefix of length 'n'. Then,
** rotate x n == BA. But BA == (A^r . B^r)^r.
*/
LUA_API void lua_rotate (lua_State *L, int idx, int n) {
StkId p, t, m;
lua_lock(L);
t = L->top.p - 1; /* end of stack segment being rotated */
p = index2stack(L, idx); /* start of segment */
api_check(L, (n >= 0 ? n : -n) <= (t - p + 1), "invalid 'n'");
m = (n >= 0 ? t - n : p - n - 1); /* end of prefix */
reverse(L, p, m); /* reverse the prefix with length 'n' */
reverse(L, m + 1, t); /* reverse the suffix */
reverse(L, p, t); /* reverse the entire segment */
lua_unlock(L);
}