[LeetCode]--189. Rotate Array

Rotate an array of n elements to the right by k steps.

For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

[show hint]

Hint:
Could you do it in-place with O(1) extra space?
Related problem: Reverse Words in a String II

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

我想到的就是反转，但是开始写的反转用了惯性思维，用下标控制，很不理想，花费大量时间还弄不精确，后来看了个湄公河，嘿嘿，一下反应到用m，n分别是起始位置和结束位置，然后一直到m

public void rotate(int[] nums, int k) {
        k = k % nums.length;
        if (k >= 1 && nums.length > k) {
            reverseArr1(nums, 0, nums.length - 1);
            reverseArr1(nums, 0, k - 1);
            reverseArr1(nums, k, nums.length - 1);
        }
    }
/**
     * 
     * @param nums
     * @param m
     *            要交换的起始位置，按照数组下标0开始
     * @param n
     *            要交换的结束位置，按照数组下标0开始
     */
    public void reverseArr1(int[] nums, int m, int n) {
        int temp = 0;
        while (m < n) {
            temp = nums[m];
            nums[m] = nums[n];
            nums[n] = temp;
            m++;
            n--;
        }
    }

这也应该算是最佳解了。给大家看看最开始写的逆转算法。蠢死了

/**
     * @param nums
     * @param m
     *            起始位置1开始
     * @param n
     *            结束位置
     */
    public void reverseArr(int[] nums, int m, int n) {
        int temp = 0, len = n - 1;
        for (int i = m - 1; i < ((n - m) / 2 + m); i++) {
            temp = nums[i];
            nums[i] = nums[len - i];
            nums[len - i] = temp;
        }
    }

还有四个解法，明天早上更新吧。

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