leetcode189.轮转数组

⭐️ 题目描述

leetcode链接：轮转数组

⭕️ C代码实现：

/*
    [1,2,3,4,5,6,7], k = 3 , numsSize = 7
    思路：
       1. 逆置 (0) - (numsSize - k - 1) 下标的元素
            - [4 , 3 , 2 , 1 , 5 , 6 , 7]
       2. 逆置 (numsSize - k) - (numsSize - 1) 下标的元素
            - [3 , 2 , 1 , 7 , 6 , 5 , 4]
       3. 整体逆置 (0) - (numsSize - 1)
            - [4 , 5 , 6 , 7 , 1 , 2 , 3]
*/
void reverse (int * range , int left , int right) {
    while (left < right) {
        int temp = range[left];
        range[left] = range[right];
        range[right] = temp;

        left++;
        right--;
    }
}

void rotate(int* nums, int numsSize, int k){
    // 防止下面操作数组逆置越界
    k %= numsSize;

    reverse(nums , 0 , numsSize - k - 1);
    reverse(nums , numsSize - k , numsSize - 1);
    reverse(nums , 0 , numsSize - 1);
}