135 Candy 分发糖果
Description:
There are N children standing in a line. Each child is assigned a rating value.
You are giving candies to these children subjected to the following requirements:
Each child must have at least one candy.
Children with a higher rating get more candies than their neighbors.
What is the minimum candies you must give?
Example:
Example 1:
Input: [1,0,2]
Output: 5
Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
Example 2:
Input: [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
The third child gets 1 candy because it satisfies the above two conditions.
题目描述:
老师想给孩子们分发糖果，有 N 个孩子站成了一条直线，老师会根据每个孩子的表现，预先给他们评分。
你需要按照以下要求，帮助老师给这些孩子分发糖果：
每个孩子至少分配到 1 个糖果。
相邻的孩子中，评分高的孩子必须获得更多的糖果。
那么这样下来，老师至少需要准备多少颗糖果呢？
示例 :
示例 1:
输入: [1,0,2]
输出: 5
解释: 你可以分别给这三个孩子分发 2、1、2 颗糖果。
示例 2:
输入: [1,2,2]
输出: 4
解释: 你可以分别给这三个孩子分发 1、2、1 颗糖果。
第三个孩子只得到 1 颗糖果，这已满足上述两个条件。
思路:
- 给每个人初始化 1个糖果
先正向遍历数组, 记录比前面大的评分的孩子的糖果为前一人加 1
再反向遍历数组, 记录比后面大的评分的孩子的糖果为后一人加 1
时间复杂度O(n), 空间复杂度O(n) - 由于分配糖果的形式一定是从 1 -> n或者从 m -> 1, 所以我们可以记录上坡的长度和下坡的次数及峰顶和谷底的糖果数, 用等差数列求和求出各个坡的糖果数
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int candy(vector& ratings)
{
vector nums(ratings.size(), 1);
for (int i = 1; i < ratings.size(); i++) if (ratings[i] > ratings[i - 1]) nums[i] = nums[i - 1] + 1;
for (int i = ratings.size() - 2; i > -1; i--) if (ratings[i] > ratings[i + 1] and nums[i] <= nums[i + 1]) nums[i] = nums[i + 1] + 1;
return accumulate(nums.begin(), nums.end(), 0);
}
};
Java:
class Solution {
public int candy(int[] ratings) {
if (ratings.length <= 1) return ratings.length;
int result = 0, up = 0, down = 0, oldSlope = 0;
for (int i = 1; i < ratings.length; i++) {
int newSlope = (ratings[i] > ratings[i - 1]) ? 1 : (ratings[i] < ratings[i - 1] ? -1 : 0);
if ((oldSlope > 0 && newSlope == 0) || (oldSlope < 0 && newSlope >= 0)) {
result += count(up) + count(down) + Math.max(up, down);
up = 0;
down = 0;
}
if (newSlope > 0) up++;
if (newSlope < 0) down++;
if (newSlope == 0) result++;
oldSlope = newSlope;
}
result += count(up) + count(down) + Math.max(up, down) + 1;
return result;
}
private int count(int a) {
return ((a * (a + 1)) >> 1);
}
}
Python:
class Solution:
def candy(self, ratings: List[int]) -> int:
candies = [1] * len(ratings)
for i in range(1, len(ratings)):
if ratings[i] > ratings[i - 1]:
candies[i] = candies[i - 1] + 1
for i in range(len(candies) - 2, -1, -1):
if ratings[i] > ratings[i + 1] and candies[i] <= candies[i + 1]:
candies[i] = candies[i + 1] + 1
return sum(candies)