LeetCode 189. Rotate Array

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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.


Have two methods there, nothing special

#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

void rotate(vector<int>& nums, int k) {
    k = k % nums.size();  // this is the only place needs attention.
    reverse(nums.begin(), nums.end());
    reverse(nums.begin(), nums.begin() + k);  // in-place
    reverse(nums.begin() + k, nums.end());
}

void rotateII(vector<int>& nums, int k) {
    int size = nums.size();
    for(int i = 0; i < size - k; ++i) {
        nums.push_back(nums[i]);
    }
    vector<int> newnums(nums.begin() + size - k, nums.end());
    nums = newnums;
}

int main(void) {
    vector<int> nums;
    nums.push_back(1);
    nums.push_back(2);
    nums.push_back(3);
    nums.push_back(4);
    nums.push_back(5);
    nums.push_back(6);
    nums.push_back(7);
    rotateII(nums, 3);
    for(int i = 0; i < nums.size(); ++i) {
        cout << nums[i] << endl;
    }
    cout << endl;
}


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leetcode 189. Rotate Array-旋转数组

原题链接:189. Rotate Array 【思路1-Java、Python】时间复杂度O(n),空间复杂度O(1) 以[1,2,3,4,5,6,7], k=3 为例,经过3次翻转,就能得到结果: reverse(nums, 0, nums.length - 1); //[7,6,5,4,3,2,1] reverse(nums, 0, k - 1);

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LeetCode(189) Rotate Array

题目如下: 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: Do it in place.Try to come up as many

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LeetCode Array 189 Rotate Array

189. Rotate Array Given an array, rotate the array to the right by k steps, where k is non-negative. Example 1: Input: [1,2,3,4,5,6,7] and k = 3 Output: [5,6,7,1,2,3,4] Explanation: rotate 1 steps to...

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