文章目录
1103. Distribute Candies to People
Problem Description
We distribute some number of candies, to a row of n = num_people people in the following way:
We then give 1 candy to the first person, 2 candies to the second person, and so on until we give n candies to the last person.
Then, we go back to the start of the row, giving n + 1 candies to the first person, n + 2 candies to the second person, and so on until we give 2 * n candies to the last person.
This process repeats (with us giving one more candy each time, and moving to the start of the row after we reach the end) until we run out of candies. The last person will receive all of our remaining candies (not necessarily one more than the previous gift).
Return an array (of length num_people and sum candies) that represents the final distribution of candies.
Example 1:
Input: candies = 7, num_people = 4
Output: [1,2,3,1]
Explanation:
On the first turn, ans[0] += 1, and the array is [1,0,0,0].
On the second turn, ans[1] += 2, and the array is [1,2,0,0].
On the third turn, ans[2] += 3, and the array is [1,2,3,0].
On the fourth turn, ans[3] += 1 (because there is only one candy left), and the final array is [1,2,3,1].
Example 2:
Input: candies = 10, num_people = 3
Output: [5,2,3]
Explanation:
On the first turn, ans[0] += 1, and the array is [1,0,0].
On the second turn, ans[1] += 2, and the array is [1,2,0].
On the third turn, ans[2] += 3, and the array is [1,2,3].
On the fourth turn, ans[0] += 4, and the final array is [5,2,3].
Constraints:
- 1 <= candies <= 10^9
- 1 <= num_people <= 1000
Solution Method
int* distributeCandies(int candies, int num_people, int* returnSize)
{
int turns = 0;
int * resArr = (int *) malloc (sizeof(int) * num_people);
memset(resArr, 0, sizeof(int) * num_people);
*returnSize = num_people;
if (num_people == 0)
return NULL;
for (int i = 0; candies > 0; i ++)
{
if (i == num_people)
{
i = 0;
turns ++;
}
// 当candies不够时，将candies发完
resArr[i] += candies >(turns * num_people + i+1)? (turns * num_people + i+1): candies;
candies -= turns * num_people + i+1;
}
return resArr;
}
61. Rotate List
Problem Description
Given a linked list, rotate the list to the right by k places, where k is non-negative.
Example 1:
Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL
Example 2:
Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL
Solution Method
基本思路：将链表变成一个循环链表，在合适的位置断开就好了。
struct ListNode* rotateRight(struct ListNode* head, int k)
{
struct ListNode * p = head, * q = head, *t = NULL;
int count = 0, sum = 0;
if (p == NULL)
return NULL;
// 找到倒数第k个元素
while (p != NULL)
{
sum ++;
p=p->next;
}
p = head;
k = k >= sum ? k % sum : k; // 确保k < sum
if (k == 0)
return head;
while (p->next != NULL)
{
if (count < k)
p=p->next;
else
{
p = p->next;
q = q->next;
}
count ++;
}
// 断开
t = q->next;
q->next = NULL;
p->next = head;
return t;
}
