LeetCode -- Candy

题目：

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?

就是给定1个rating数组，表示每个娃所得的评分。现在要求：
1.给每个娃至少分1个candy
2.分高的娃要比邻居所得的candy多
3.尽量少分candy

思路：
此题采用贪心法。

贪心策略：
1. 先分给谁？ 从得分最少的娃开始分，因此不妨构造child对象，对得分排序，以此存入child。
2. 分多少？ 由于是按照得分从低到高分candy的。 如果邻居已经有人得了candy，那么就分别将当前娃的得分与左右两边娃进行对比：
a.如果当前得分与左或右相等，只给1个
b.如果当前得分大于左或右，需要给对方所得的candy数+1个
综合a,b，取最大的那个。

如果两边没人得candy，只给当前娃1个。

实现代码：

class Child
{
public Child(int i , int v){
index = i;
rating = v;
}

public int index;
public int rating;
}

public int Candy(int[] ratings)
{
if(ratings == null || ratings.Length == 0){
return 0;
}

if(ratings.Length == 1){
return 1;
}

var arr = new List<Child>();
var candies = new int[ratings.Length];
for(var i = 0 ;i < ratings.Length; i++){
candies[i] = 0;
arr.Add(new Child(i, ratings[i]));
}

arr = arr.OrderBy(x=>x.rating).ToList();

for(var i = 0 ;i < arr.Count ; i++){
var index = arr[i].index;
if(candies[index] != 0){
continue;
}

if(index == 0){
candies[index] = ratings[index] == ratings[index + 1]  ? 1 : candies[index + 1] + 1;
}
else if(index == candies.Length - 1){
candies[index] = ratings[index] == ratings[index - 1]  ? 1 : candies[index - 1] + 1;
}
else{
var left = ratings[index-1] == ratings[index] ? 1 : candies[index - 1] + 1;
var right = ratings[index+1] == ratings[index] ? 1 : candies[index + 1] + 1;

candies[index] = Math.Max(left , right);
}

}

var s = 0;
for(var i = 0;i < candies.Length; i++){
s += candies[i];
}

return s;
}