用于数组旋转的块交换算法的 Javascript 程序
原文:https://www . geesforgeks . org/JavaScript-程序换块-交换-算法换数组-旋转/
编写一个函数 rotate(ar[]，d，n)，将大小为 n 的 arr[]旋转 d 个元素。
将上面的数组旋转 2 将构成数组
算法:
Initialize A = arr[0..d-1] and B = arr[d..n-1]
1) Do following until size of A is equal to size of B
a) If A is shorter, divide B into Bl and Br such that Br is of same
length as A. Swap A and Br to change ABlBr into BrBlA. Now A
is at its final place, so recur on pieces of B.
b) If A is longer, divide A into Al and Ar such that Al is of same
length as B Swap Al and B to change AlArB into BArAl. Now B
is at its final place, so recur on pieces of A.
2) Finally when A and B are of equal size, block swap them.
递归实现:
java 描述语言
<script>
let leftRotate = (arr, d, n) =>{
/* Return If number of elements to be rotated
is zero or equal to array size */
if(d == 0 || d == n)
return;
/*If number of elements to be rotated
is exactly half of array size */
if(n - d == d) {
arr = swap(arr, 0, n - d, d);
return;
}
/* If A is shorter*/
if(d < n - d) {
arr = swap(arr, 0, n - d, d);
leftRotate(arr, d, n - d);
}
else{
/* If B is shorter*/
arr = swap(arr, 0, d, n - d);
/*This is tricky*/
leftRotate(arr + n - d, 2 * d - n, d);
}
}
/*UTILITY FUNCTIONS*/
/* function to print an array */
let printArray = (arr, size) =>{
ans = ''
for(let i = 0; i < size; i++)
ans += arr[i]+" ";
document.write(ans)
}
/*This function swaps d elements
starting at index fi
with d elements starting at index si */
let swap = (arr, fi, si, d) =>{
for(let i = 0; i < d; i++) {
let temp = arr[fi + i];
arr[fi + i] = arr[si + i];
arr[si + i] = temp;
}
return arr
}
// Driver Code
arr = [1, 2, 3, 4, 5, 6, 7];
leftRotate(arr, 2, 7);
printArray(arr, 7);
</script>
输出:
3 5 4 6 7 1 2
迭代实现: 这里是同样算法的迭代实现。这里使用了相同的实用函数 swap()。
java 描述语言
<script>
// JavaScript code for above implementation
function leftRotate(arr, d, n)
{
if(d == 0 || d == n)
return;
let i = d;
let j = n - d;
while (i != j)
{
if(i < j)
{
// A is shorter
arr = swap(arr, d - i, d + j - i, i);
j -= i;
}
else{ // B is shorter
arr = swap(arr, d - i, d, j);
i -= j;
}
}
arr = swap(arr, d - i, d, i);
// This code is contributed by rohitsingh04052.
}
</script>
时间复杂度: O(n) 其他阵轮换法请看以下帖子: https://www.geeksforgeeks.org/array-rotation/ https://www . geeksforgeeks . org/程序换阵-轮换-续-反转-算法/
参考文献: 【http://www.cs.bell-labs.com/cm/cs/pearls/s02b.pdf】 如发现以上程序/算法有任何 bug 或想分享任何关于区块交换算法的补充信息，请写评论。
更多详情请参考数组旋转的块交换算法的完整文章！
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