Candy

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?

贪心法

复杂度

时间 O(N) 空间 O(N)

思路

典型的贪心法，如果一个孩子比另一个孩子的分高，我们只多给1块糖。我们可以先从左往右遍历，确保每个孩子根他左边的孩子相比，如果分高，则糖要多1个，如果分比左边低，就只给一颗。然后我们再从右往左遍历，确保每个孩子跟他右边的孩子相比，如果分高则糖至少多1个（这里至少多1个的意思是，我们要取当前孩子手里糖的数量，和其右边孩子糖的数量加1，两者的较大值）。

代码

public class Solution {
    public int candy(int[] ratings) {
        if(ratings.length <= 1){
            return ratings.length;
        }
        int[] candies = new int[ratings.length];
        candies[0] = 1;
        // 先从左往右分糖，分数较高的多拿一颗糖，分数较少的只拿一颗糖
        for(int i = 1; i < ratings.length; i++){
            if(ratings[i] > ratings[i - 1]){
                candies[i] = candies[i - 1] + 1;
            } else {
                candies[i] = 1;
            }
        }
        int sum = candies[candies.length - 1];
        // 再从右往左继续分糖，分数较高的确保比右边多一颗就行了
        for(int i = ratings.length - 2; i >= 0; i--){
            if(ratings[i] > ratings[i + 1]){
                candies[i] = Math.max(candies[i + 1] + 1, candies[i]);
            }
            sum += candies[i];
        }
        return sum; 
    }
}