#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
//分糖果的问题
/*
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.
*/
/*
解题思路:
遍历两边,首先每个人得一块糖,第一遍从左到右,若当前点比前一个点高就比前者多一块。
这样保证了在一个方向上满足了要求。第二遍从右往左,若左右两点,左侧高于右侧,但
左侧的糖果数不多于右侧,则左侧糖果数等于右侧糖果数+1,这就保证了另一个方向上满足要求。
最后将各个位置的糖果数累加起来就可以了。
*/
int candyCount(vector<int>&rating) {
int res = 0;
//孩子总数
int n = rating.size();
//糖果集合
vector<int> candy(n, 1);
//从左往右遍历
for (int i = 0;i < n - 1;i++) {
if (rating[i + 1] > rating[i])candy[i + 1] = candy[i] + 1;
}
//从右往左
for (int i = n - 1;i > 0;i--) {
if (rating[i - 1] > rating[i] && candy[i - 1] <= candy[i])candy[i - 1] = candy[i] + 1;
}
//累加结果
for (auto a : candy) {
res += a;
}
return res;
}
//测试函数
int main() {
vector<int> rating{1,3,2,1,4,5,2};
cout << candyCount(rating) << endl;
return 0;
}