Candy

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?

題目說N個孩子排成一行，每個孩子被賦予一個rating，要求給candy給孩子滿足每個孩子都至少有一個candy同時擁有更高rating

的孩子比他的鄰居有更多的candy。

思路1：轉爲有向圖，用拓撲排序。

拓撲排序用到了每個節點的入度（有多少個節點指向該節點）、從該節點出發直接到達的節點。

步驟：

1. 找到入度爲0的節點，加入隊列中；
2. 隊列頭部取出元素，累加candy數；
3. 從該節點出發的節點的入度都要減1，如果達到入度爲0，則該節點的candy數爲當前節點candy數加1，,加入隊列中。

<span style="font-size:14px;">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;
    }
};</span>

思路2：遞歸方法

遞歸是個很強大的方法，這裏也可以用遞歸來求出，我們要找出每個位置的candy數，這個candy數取決於該位置與

相鄰位置的ratings大小情況。遞歸就是假設鄰居位置的candy數已經求好了，然後計算該位置的candy數，要注意好

邊界條件的判斷。

<span style="font-size:14px;">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;
    }
};</span>

上面的代碼邏輯有點混亂，可以精簡成如下：

<span style="font-size:14px;">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, ratings);
        }
        
        return ans;
    }
private:
    int get_candy_num(vector<int> &candy_num, size_t k, vector<int> &ratings) {
        if (candy_num[k]) return candy_num[k];

        candy_num[k] = 1;
        
        if (k > 0 && ratings[k] > ratings[k - 1]) {
            candy_num[k] = max(candy_num[k], get_candy_num(candy_num, k - 1, ratings) + 1);
        }
        if (k + 1 < candy_num.size() && ratings[k] > ratings[k + 1]) {
            candy_num[k] = max(candy_num[k], get_candy_num(candy_num, k + 1, ratings) + 1);
        }
        
        return candy_num[k];
    }
};</span>