[leetcode] Rotate Array

最新推荐文章于 2024-01-03 12:46:00 发布

原创最新推荐文章于 2024-01-03 12:46:00 发布 · 350 阅读

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本文介绍了一种将数组元素向右旋转k步的算法实现。通过两种不同的方法实现了这一操作：一种是逐个移动元素，另一种是利用异或操作进行原地交换，避免了额外的空间消耗。

From :

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].

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        if(k==0) return;
        k= k % nums.size();
        while(k--) {
            nums.insert(nums.begin(), nums.back());
            nums.pop_back();
        }
    }
};

class Solution {
public:
    void rotate(vector<int>& nums, int start, int end) {
        while(start < end) {
            nums[start] ^= nums[end];
            nums[end] ^= nums[start];
            nums[start] ^= nums[end];
            ++start;
            --end;
        }
    }
    void rotate(vector<int>& nums, int k) {
        k = k%nums.size();
        rotate(nums, 0, nums.size()-1);
        rotate(nums, 0, k-1);
        rotate(nums, k, nums.size()-1);
    }
};

public class Solution {
    public void rotate(int[] nums, int k) {
		int n;
		if (nums == null || (n = nums.length) == 0) {
			return;
		}
		k = k % nums.length;
		rotate(nums, 0, n - 1);
		rotate(nums, 0, k - 1);
		rotate(nums, k, n - 1);
	}

	private void rotate(int[] nums, int start, int end) {
		while (start < end) {
			nums[start] ^= nums[end];
			nums[end] ^= nums[start];
			nums[start] ^= nums[end];
			++start;
			--end;
		}
	}
}