[LeetCode] Rotate Array 旋转数组

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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]

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

新题抢先刷，这道题标为Easy，应该不是很难，题目明确要求了空间复杂度为O(1)，那么肯定不能建立新的数组，所以只能覆盖原数组的值，仔细观察题目中给的那个例子，k = 3说是移动了三步，怎么移的呢，原来是每次把最末尾的数移动到开头，移三次就得到最终结果。代码如下：

class Solution {
public:
    void rotate(int nums[], int n, int k) {
        for (int i = 0; i < k; ++i) {
            moveRight(nums, n);
        }        
    }
    void moveRight(int nums[], int n) {
        if (n == 0) return;
        int last = nums[n - 1];
        for (int i = n - 1; i >= 1; --i) nums[i] = nums[i - 1];
        nums[0] = last;
    }
};

题目说至少有三种解法，暂时只想到一种，容我慢慢想想~~

[LeetCode] Rotate Array 旋转数组

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原文地址：http://www.cnblogs.com/grandyang/p/4298711.html