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Candy

2024-07-19 05:26:39   226人阅读

Candy

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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?

题目说N个孩子排成一行，每个孩子被赋予一个rating，要求给candy给孩子满足每个孩子都至少有一个candy同时拥有更高rating

的孩子比他的邻居有更多的candy。

思路1：转为有向图，用拓扑排序。

拓扑排序用到了每个节点的入度（有多少个节点指向该节点）、从该节点出发直接到达的节点。

步骤：

  1. 找到入度为0的节点，加入队列中；
  2. 队列头部取出元素，累加candy数；
  3. 从该节点出发的节点的入度都要减1，如果达到入度为0，则该节点的candy数为当前节点candy数加1，,加入队列中。
class Solution {
public:
    int candy(vector<int> &ratings) {
        size_t len = ratings.size();
        
        if (len == 1) {
            return 1;
        }
        vector<int> ins(len, 0);
        vector<vector<int> > outs(len, vector<int>());
        
        for (int i = 0; i < len; ++i) {
            if (i - 1 >= 0 && ratings[i - 1] > ratings[i]) {
                ++ins[i - 1];
                outs[i].push_back(i - 1);
            }
            if (i + 1 < len && ratings[i + 1] > ratings[i]) {
                ++ins[i + 1];
                outs[i].push_back(i + 1);
            }
        }
        
        queue<pair<int, int> > q;
        for (int i = 0; i < len; ++i) {
            if (!ins[i]) {
                q.push(make_pair(i, 1));
            }
        }
        
        int ans = 0;
        while (!q.empty()) {
            pair<int, int> front = q.front();
            q.pop();
            
            ans += front.second;
            for (int i = 0; i < outs[front.first].size(); ++i) {
                --ins[outs[front.first][i]];
                if (!ins[outs[front.first][i]]) {
                    q.push(make_pair(outs[front.first][i], front.second + 1));
                }
            }
        }
        
        return ans;
    }
};

思路2：递归方法
递归是个很强大的方法，这里也可以用递归来求出，我们要找出每个位置的candy数，这个candy数取决于该位置与
相邻位置的ratings大小情况。递归就是假设邻居位置的candy数已经求好了，然后计算该位置的candy数，要注意好
边界条件的判断。
class Solution {
public:
    int candy(vector<int> &ratings) {
        size_t len = ratings.size();
        if (len == 1) {
            return 1;
        }
        vector<int> candy_num(len, 0);
        
        int ans = 0;
        for (size_t i = 0; i < len; ++i) {
            ans += get_candy_num(candy_num, i, len, ratings);
        }
        
        return ans;
    }
private:
    int get_candy_num(vector<int> &candy_num, size_t k, size_t len, vector<int> &ratings) {
        if (candy_num[k]) return candy_num[k];

        if (k == 0) {
            if (ratings[k + 1] >= ratings[k]) {
                return candy_num[k] = 1;
            } else {
                return candy_num[k] = get_candy_num(candy_num, k + 1, len, ratings) + 1;
            }
        } else if (k == len - 1) {
            if (ratings[k - 1] >= ratings[k]) {
                return candy_num[k] = 1;
            } else {
                return candy_num[k] = get_candy_num(candy_num, k - 1, len, ratings) + 1;
            }
        } else {
            if (ratings[k - 1] >= ratings[k] && ratings[k + 1] >= ratings[k]) {
                return candy_num[k] = 1;
            } else if (ratings[k] > ratings[k - 1] && ratings[k] > ratings[k + 1]) {
                int left = get_candy_num(candy_num, k - 1, len, ratings);
                int right = get_candy_num(candy_num, k + 1, len, ratings);
                left < right ? candy_num[k] = right + 1 : candy_num[k] = left + 1;
                return candy_num[k];
            } else if (ratings[k] > ratings[k - 1]) {
                return candy_num[k] = get_candy_num(candy_num, k - 1, len, ratings) + 1;
            } else if (ratings[k] > ratings[k + 1]) {
                return candy_num[k] = get_candy_num(candy_num, k + 1, len, ratings) + 1;
            } 
        }
        
        return 1;
    }
};


Candy


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