[leetcode]Candy @ Python
2014-09-14 11:06
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原题地址:https://oj.leetcode.com/problems/candy/
题意:
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?
解题思路:求最吝啬的糖果分发数。先从前到后扫描一遍数组,如果序列递增,就+1;然后从后到前扫描一遍数组,序列递增,+1。保证最低谷(ratings最小)永远是1就可以了。
代码: 时间复杂度O(n),空间复杂度O(n)
题意:
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?
解题思路:求最吝啬的糖果分发数。先从前到后扫描一遍数组,如果序列递增,就+1;然后从后到前扫描一遍数组,序列递增,+1。保证最低谷(ratings最小)永远是1就可以了。
代码: 时间复杂度O(n),空间复杂度O(n)
class Solution: # @param ratings, a list of integer # @return an integer def candy(self, ratings): candyNum = [1] * len(ratings) for i in range(1, len(ratings)): if ratings[i - 1] < ratings[i]: candyNum[i] = candyNum[i - 1] + 1 for i in range(len(ratings) - 2, -1, -1): if ratings[i] > ratings[i + 1] and candyNum[i] <= candyNum[i + 1]: candyNum[i] = candyNum[i + 1] + 1 return sum(candyNum)
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