[leetcode] 135. Candy 解题报告

题目链接: https://leetcode.com/problems/candy/

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?

思路: 自己想了一个思路比较麻烦,利用栈来维护一个递减序列. 

写完之后又看了一下别人怎么写的, 发现只需要从左往右更新一遍计数, 再从右往左扫描更新一遍计数就可以, 这种思路比较容易理解.

两种代码时间复杂度都是O(n), 但是第二种要更简洁明了.

代码如下:

class Solution {
public:
    int candy(vector<int>& ratings) {
        if(ratings.size() ==0) return 0;
        stack<int> st;
        int sum = 1, num = 1, gap = 0;
        st.push(ratings[0]);
        for(int i =1; i<ratings.size(); i++)
        {
            if(ratings[i] > st.top())
            {
                while(!st.empty()) st.pop();
                num++;
            }
            else if(ratings[i] < st.top())
            {
                sum += st.size();
                if(num >1) gap = num - 1;
                if(gap >= 1) sum--, gap--;
                num = 1;
            }
            else if(ratings[i] == st.top())
            {
                while(!st.empty()) st.pop();
                gap = 0;
                num = 1;
            }
            st.push(ratings[i]);
            sum += num;
        }
        return sum;
    }
};

class Solution {
public:
    int candy(vector<int>& ratings) {
        int sum = 0;
        vector<int> vec(ratings.size(), 1);
        for(int i = 1; i< ratings.size(); i++)
            if(ratings[i] > ratings[i-1]) vec[i]=vec[i-1] +1;
        for(int i = ratings.size()-2; i>=0; i--)
            if(ratings[i] > ratings[i+1] && vec[i]<=vec[i+1])
                vec[i] = vec[i+1] + 1;
        for(auto val: vec) sum += val;
        return sum;
    }
};

第二种参考: https://leetcode.com/discuss/92299/simple-c-dp-solution-36ms-beats-88%25-submissions