135. Candy Leetcode Python

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本文介绍了一种有效解决糖果分配问题的方法，确保每个孩子至少得到一颗糖果，并且评分高的孩子比邻居得到更多糖果。通过两次遍历，从左到右再从右到左，实现了O(n)的时间复杂度。

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?

这道题需要两次循环，一次从左边往右边，一次从右边往左边。每次扫描维护最小的满足条件的数组。

时间复杂度为O(n)空间复杂度也为O(n)

We need to iterate twice, first time from left to right, second time from right to left and maintain the smallest value.

The time complexity is O(n) and space complexity is O(n).

class Solution:
    # @param ratings, a list of integer
    # @return an integer
    def candy(self, ratings):
        if len(ratings)==0:
            return 0
        candies=[1 for index in range(len(ratings))]
        for index in range(1,len(ratings)):
            if ratings[index]>ratings[index-1]:
                candies[index]=candies[index-1]+1
        for index in reversed(range(1,len(ratings))):
            if ratings[index]<ratings[index-1]:
                candies[index-1]=max(candies[index]+1,candies[index-1])
        return sum(candies)