189-旋转数组

LeetCode 189. Rotate Array

LeetCode-189

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

method 1: 额外的数组

把nums[i]赋值到移动后的位置，注意取模

void rotate(vector<int>& nums, int k) {
    int n = nums.size();
    vector<int> res(n);
    for (int i = 0; i < n; i++) {
        res[(i + k) % n] = nums[i];
    }
    nums.swap(res);
}

时间复杂度：$O(n)$
空间复杂度：$O(n)$

method 2: 原地翻转

void swap(vector<int>& nums, int i, int j) {
    while (i < j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
        i++, j--;
    }
}
void rotate(vector<int>& nums, int k) {
    int n = nums.size();
    k %= n;
    swap(nums, 0, n - 1);   // 1
    swap(nums, 0, k - 1);   // 2
    swap(nums, k, n - 1);   // 3
}