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Algorithm

135. Candy

Description

There are n children standing in a line. Each child is assigned a rating value given in the integer array ratings.

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. Return the minimum number of candies you need to have to distribute the candies to the children.

Example 1:

Input: ratings = [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: ratings = [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.

Constraints:

  • n == ratings.length
  • 1 <= n <= 2 * 104
  • 0 <= ratings[i] <= 2 * 104

Solution

class Solution {
    public int candy(int[] ratings) {
        int len=ratings.length;
        int [] candy=new int[len];
        // 1. 初始化糖果的数量为1
        for(int i=0;i<len;i++){
            candy[i]=1;
        }
        // 2. 从前往后:如果第i个小孩的等级>前面i-1个小孩的等级,那么糖果数+1
        for(int i=1;i<len;i++){
            if(ratings[i]>ratings[i-1]){
                candy[i]=candy[i-1]+1;
            }
        }
        // 3. 从后往前: 如果第i个小孩的等级>后面i+1个小孩的等级且糖果数少,那么糖果数+1
        for(int i=len-2;i>=0;i--){
            if((ratings[i]>ratings[i+1])&&(candy[i]<=candy[i+1])){
                candy[i]=candy[i+1]+1;   
            }
        }
        int num=0;
        for(int i=0;i<len;i++){
            num+=candy[i];
        }
        return num;
    }
}

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