[Java][Leetcode simple] 189. 轮转数组

借助辅助数组

  借助一个辅助数组tmp保存后面k个元素然后逆序循环，使用数组前面n-k个元素覆盖最后到后面最后把前k个元素从tmp中拿回来

 public void rotate(int[] nums, int k) {int len = nums.length;k = k % len;int[] Ra = new int[len];int cnt = 0;for (int i = len-k ; i < len; i++) {Ra[cnt++] = nums[i];}for (int i = len-1; i >=0 ; i--) {if(i>=k){nums[i] = nums[i-k];}else{nums[i] = Ra[i];}}}

借助数组旋转

先整体旋转
然后0~k-1旋转
最后k-n-1旋转，即可得到目标数组

    public void rotate2(int[] nums, int k) {int len = nums.length;k = k % len;reverse(nums, 0, len-1);reverse(nums, 0, k-1);reverse(nums, k, len-1);}public void reverse(int[] nums, int start, int end) {while (start < end) {int temp = nums[start];nums[start] = nums[end];nums[end] = temp;start++;end--;}}