leetcode 135 Candy

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本文深入探讨了Candy分配算法的实现，旨在确保每位儿童基于其评分获得的糖果数量不仅公平，而且达到最小总消耗。通过两次遍历，分别从前向后和从后向前比较儿童评分，动态调整糖果数量，确保高评分儿童比邻居获得更多糖果，同时维持最低糖果使用量。

135. 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?

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.

Solutions:

if __name__== "__main__":
    def candy(ratings):
        lenr = len(ratings)                //lenr为ratings的长度
        if lenr<=1: return lenr
        dp = [1 for _ in range(lenr)]      // 建立长为lenr，每个位置为1的数组dp
        for i in range(1, lenr):           //从前往后，从后往前遍历两次，取最大值
            if ratings[i] > ratings[i-1] :
                dp[i] = dp[i-1]+1
        for j in range(lenr-1, 0, -1):
            if ratings[j] < ratings[j-1]:
                dp[j-1] = max(dp[j-1], dp[j]+1)
        return sum( dp )
    print(candy([1,0,2]))