leetcode candy

最新推荐文章于 2019-10-23 11:09:46 发布

原创最新推荐文章于 2019-10-23 11:09:46 发布 · 216 阅读

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本文介绍了一种高效的算法，用于解决儿童按评分公平分配糖果的问题。确保每个孩子至少获得一颗糖果，并且评分较高的孩子得到的糖果数量多于邻居。通过双向遍历数组的方法，实现了最小糖果数目的计算。

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?


class Solution {
public:
    int candy(vector<int> &ratings) 
    {
        vector<int> res(ratings.size(), 1);
        for(int i=0;i<ratings.size()-1;++i)
        {
            if (ratings[i+1] > ratings[i])
                res[i+1] = res[i]+1;
        }
        for(int i=ratings.size()-1;i>=0;--i)
        {
            if (ratings[i-1] > ratings[i])
                res[i-1] = res[i]+1;
        }
        return accumulate(res.begin(), res.end(), 0);
    }
};

测试

#include<iostream>
#include<vector>
#include <numeric>
using namespace std;
class Solution {
public:
    int candy(vector<int> &ratings) 
    {
        vector<int> res(ratings.size(), 1);
        for(int i=0;i<ratings.size()-1;++i)
        {
            if (ratings[i+1] > ratings[i])
                res[i+1] = res[i]+1;
        }
        for(int i=ratings.size()-1;i>=0;--i)
        {
            if (ratings[i-1] > ratings[i]&&res[i-1]<=res[i])
                res[i-1] = res[i]+1;
        }
        return accumulate(res.begin(), res.end(), 0);
    }
};

int main()
{
    Solution s;
    int A[] = {4,2,3,4,1};
    vector<int> vec(A, A+5);
    cout<<s.candy(vec)<<endl;
}