189. 旋转数组

题意：

解题思路：

1. 第一种方法是，先将k后的数字放入临时数组，再将k前的数字追加，最后copy到原数组
2. 第二种方法是用递归，先整体翻转，再以k为中心，翻转2个局部

PHP代码实现：

class Solution {
    /**
     * @param Integer[] $nums
     * @param Integer $k
     * @return NULL
     */
    function rotate(&$nums, $k) {
        for ($i = 0; $i < count($nums); $i++) {
            $a[($i + $k) % count($nums)] = $nums[$i];
        }
        for ($i = 0; $i < count($nums); $i ++) {
            $nums[$i] = $a[$i];
        }
        return $nums;
    }
    function rotateByCopy(&$nums, $k) {
        $n = count($nums);
        $m = $k % $n; $i = 0;
        for ($j = $n - $m; $j < $n; $j++) $t[$i++] = $nums[$j];
        for ($j = 0; $j < $n - $m; $j ++) $t[$i++] = $nums[$j];
        for ($j = 0; $j < $n; $j++) $nums[$j] = $t[$j];
    }
    function rotateBySwap(&$nums, $k) {
        if ($nums == null || count($nums) == 0 || $k <= 0) return;
        $n = count($nums); $m = $k % $n;
        $this->reverse($nums, 0, $n - 1);
        $this->reverse($nums, 0, $m - 1);
        $this->reverse($nums, $m, $n - 1);
    }
    function reverse(&$nums, $i, $j) {
        for (; $i < $j; ++$i, --$j) {
            $tmp = $nums[$i];
            $nums[$i] = $nums[$j];
            $nums[$j] = $tmp;
        }
    }
}

go代码实现：

func rotate(nums []int, k int)  {
    n := len(nums)
    m := k % n
    i := 0
    res := make([]int, len(nums))
    for j := n - m; j < n; j++ {
        res[i] = nums[j]
        i++
    }
    for j := 0; j < n - m; j++ {
        res[i] = nums[j]
        i++
    }
    for i, v := range res {
        nums[i] = v
    }
}
func rotateCopy(nums []int, k int) {
    k = len(nums) - k%len(nums)
    copy(nums, append(nums[k:], nums[0:k]...))
}
func reverse(nums []int) {  // 翻转数组
    i,j := 0, len(nums)-1
    for i < j {
        nums[i],nums[j] = nums[j],nums[i]
        i++
        j--
    }
}
func rotateBySwap(nums []int, k int)  {
    k = k % len(nums)
    reverse(nums[:len(nums)-k])
    reverse(nums[len(nums)-k:])
    reverse(nums)
}