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?
思路：neighbor=>元素值与前、后元素有关=>
第一次前向扫描：保证high rate元素比左邻居糖多；
第二次后向扫描：保证high rate元素比右邻居糖多；
class Solution {
public:
int candy(vector<int> &ratings) {
vector<int> new_ratings(ratings.size(), );
for(int i = ; i < ratings.size(); i++) //for the first time, scan from beginning
{
if(ratings[i] > ratings[i-])
{
new_ratings[i] = new_ratings[i-]+;
}
}
for(int i=ratings.size()-;i>=;i--){ //for the second time, scan from the end
if(ratings[i]>ratings[i+]&&new_ratings[i]<=new_ratings[i+]){
new_ratings[i]=new_ratings[i+]+;
}
}
int sum=;
for(int i=;i<new_ratings.size();i++){
sum+=new_ratings[i];
}
return sum;
}
};