Leetcode:Candy

Leetcode: candy

There are N children standing in a line. Each child is assigned arating value.
You are giving candies to these children subjected to the followingrequirements:
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?
思路：

因为每个人最少得到一个糖果，初始化Num_arr[]为1，表示每个人得到的起始糖果为1。

然后从起始到末尾遍历ratings数组，假如第i个的rating value比i-1的大，就将此rating value代表的Num_arr[i]变为Num_arr[i-1]+1。表示比前一个的糖果数多一个。

然后再从末尾到起始遍历ratings数组，假如第i-1个的ratingvalue比i的大，就将当前的Num_arr[i-1]与Num_arr[i]+1比较，若是Num_arr[i]+1较大，就将Num_arr[i-1]更新为Num_arr[i]+1。最终得到最小的糖果数能满足要求。

代码：

int candy(vector<int> &ratings) {
int n=ratings.size();
vector<int> Num_arr(n,1);
for(int i=1; i!=n; i++)
{
if(ratings[i]>ratings[i-1])
Num_arr[i] = Num_arr[i-1]+1;
}
for(int i=n-1; i>0; i--)
{
if(ratings[i-1]>ratings[i] && Num_arr[i-1] < Num_arr[i]+1)
Num_arr[i-1] = Num_arr[i]+1;
}
int sum=0;
for(int i=0; i!=n; i++)
sum += Num_arr[i];
return sum;
}