C++实现——小孩分糖果问题

#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;
}