【原创】leetCodeOj --- Candy 解题报告

题目地址：
https://leetcode.com/problems/candy/
题目内容：

Candy

Total Accepted: 43150 Total Submissions: 203841 Difficulty: Hard

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?

题目解析：

简单来讲，一个数组里的每个小孩都有权值，每个小孩都必须有一颗糖，而如果身边的小孩权值比自己低，那么自己就能多拿糖。

问最少能给几颗糖

容易看出，n个小孩，那么至少n颗糖，我们可以在这个基础上分配糖。

贪心算法：

从权值最小的小孩开始分起，每分到一个小孩时，看他是否比身边小孩的权值大，如果大，则比身边小孩最多糖数再多1；如果小，则不变。

为什么这个算法是对的？

0、由于权值比自己小的小孩已经处理过了，所以，后面不会出现权值比自己小的小孩获得的糖果数更新的情况，因此，自己的糖果数也不用更新。

1、由于每处理一个小孩，该小孩就不用再修改，所以从1块糖处理起时，每个小孩拿到的都是自己可以拿到的最小值。

具体代码：

class Solution(object):
def candy(self, ratings):
"""
:type ratings: List[int]
:rtype: int
"""
self.ratings = ratings
if len(ratings) == 0:
return 0
heap = list()
candies = list()
for i in range(len(ratings)):
heap.append({'index':i,'value':ratings[i]})
candies.append(1)
heap = sorted(heap, key=lambda x:x['value'])
for i in range(len(heap)):
self.get_candy(heap[i], candies)
return sum(candies)

def get_candy(self, item, candies):
index = item['index']
value = item['value']
if index - 1 >= 0:
if self.ratings[index - 1] < value:
candies[index] = candies[index - 1] + 1
if index + 1 < len(candies):
if self.ratings[index + 1] < value:
if candies[index] < candies[index + 1] + 1:
candies[index] = candies[index + 1] + 1