【LeetCode】#135分发糖果(Candy)
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：
1.每个孩子至少分配到 1 个糖果。
2.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。
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:
1.Each child must have at least one candy.
2.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.
解法
class Solution {
public int candy(int[] ratings) {
if(ratings==null || ratings.length==0)
return 0;
int[] nums = new int[ratings.length];
for(int i=0; i<ratings.length; i++){
nums[i] = 1;
}
for(int i=1; i<ratings.length; i++){
if(ratings[i]>ratings[i-1] && nums[i]<=nums[i-1]){
nums[i] = nums[i-1]+1;
}
}
for(int i=ratings.length-2; i>=0; i--){
if(ratings[i]>ratings[i+1] && nums[i]<=nums[i+1]){
nums[i] = nums[i+1]+1;
}
}
int sum = 0;
for(int i : nums){
sum += i;
}
return sum;
}
}