C++实现——小孩分糖果问题

#include 
#include 
#include 
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;
}