carloscn
commented
1 year ago Analysis
int32_t sort_even_odd(int32_t *nums, size_t nums_size)
{
int32_t ret = 0;
int32_t *a_buf = NULL;
int32_t *b_buf = NULL;
UTILS_CHECK_PTR(nums);
UTILS_CHECK_LEN(nums_size);
a_buf = (int32_t *)malloc(sizeof(int32_t) * (nums_size / 2 + 1));
UTILS_CHECK_PTR(a_buf);
b_buf = (int32_t *)malloc(sizeof(int32_t) * (nums_size / 2 + 1));
UTILS_CHECK_PTR(b_buf);
for (size_t i = 0; i < nums_size; i ++) {
if (i % 2 == 0) {
a_buf[i/2] = nums[i];
} else {
b_buf[i/2] = nums[i];
}
}
for (size_t i = 0; i < nums_size/2; i ++) {
for (size_t j = 0; j < nums_size/2 - i - 1; j ++) {
if (a_buf[j] > a_buf[j + 1]) {
utils_swap_int32(a_buf + j, a_buf + j + 1);
}
if (b_buf[j] < b_buf[j + 1]) {
utils_swap_int32(b_buf + j, b_buf + j + 1);
}
}
}
for (size_t i = 0; i < nums_size / 2; i ++) {
nums[2 * i] = a_buf[i];
nums[2 * i + 1] = b_buf[i];
}
finish:
UTILS_SAFE_FREE(a_buf);
UTILS_SAFE_FREE(b_buf);
return ret;
}
Description
You are given a 0-indexed integer array nums. Rearrange the values of nums according to the following rules:
Sort the values at odd indices of nums in non-increasing order. For example, if nums = [4,1,2,3] before this step, it becomes [4,3,2,1] after. The values at odd indices 1 and 3 are sorted in non-increasing order. Sort the values at even indices of nums in non-decreasing order. For example, if nums = [4,1,2,3] before this step, it becomes [2,1,4,3] after. The values at even indices 0 and 2 are sorted in non-decreasing order. Return the array formed after rearranging the values of nums.
Example 1:
Input: nums = [4,1,2,3] Output: [2,3,4,1] Explanation: First, we sort the values present at odd indices (1 and 3) in non-increasing order. So, nums changes from [4,1,2,3] to [4,3,2,1]. Next, we sort the values present at even indices (0 and 2) in non-decreasing order. So, nums changes from [4,1,2,3] to [2,3,4,1]. Thus, the array formed after rearranging the values is [2,3,4,1].
Example 2:
Input: nums = [2,1] Output: [2,1] Explanation: Since there is exactly one odd index and one even index, no rearrangement of values takes place. The resultant array formed is [2,1], which is the same as the initial array.
Constraints:
1 <= nums.length <= 100 1 <= nums[i] <= 100