leetcode 135. Candy

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本文介绍了一种针对儿童评分的糖果分配算法。确保每个孩子至少获得一颗糖果，并且评分高的孩子比邻居获得更多糖果。文章提供了两种算法实现，强调了代码的严谨性和正确理解题目要求的重要性。

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.

马丹，开始以为你可以自己排列这些小朋友，然后发给他们糖果，提交之后发现不对，看了测试用例之后悔悟必须按原先的排列顺序来

class Solution {
public:
	int candy(vector<int>& ratings) {
		if (ratings.empty())
			return 0;
		if (ratings == vector<int>(ratings.size(), ratings[0]))
			return ratings.size();
		map<int,int>cnt;
		for (int i = 0; i < ratings.size(); i++)
			cnt[ratings[i]]++;
		int sum = 0;
		int imax = (cnt.size() + 1) & 4 == 0 ? cnt.size() / 2 + 1 : cnt.size() / 2;
		int i = 0;
		int more2=0;
		for (map<int, int>::iterator it = cnt.begin(); it != cnt.end(); it++)
		{
			if (i < imax)
				sum += it->second;
			else
			{
				if (it->second >= 2)
				{
					more2++;
					sum += 2 * it->second - 2 * (it->second - 2);
				}
				else
					sum += 2;
			}
			i++;
		}
		if ((cnt.size() + 1) & 4 != 0 && more2 > 0)
		{
			if ((cnt.size() - 1) & 4 == 0 && more2 >= 2)
				sum -= 2;
			else
				sum -= 1;
		}
		return sum;
	}
};

下面是正确理解题意后的代码，问题不算很难，但代码严谨性要高，否则还是会出些问题的

class Solution {
public:
	int candy(vector<int>& ratings) {
		if (ratings.empty())
			return 0;
		if (ratings == vector<int>(ratings.size(), ratings[0]))
			return ratings.size();
		vector<int>cans(ratings.size());
		int k = 0;
		while (k < ratings.size())
		{
			int k1 = k;
			while (k1 + 1 < ratings.size() && ratings[k1] < ratings[k1 + 1])
				k1++;
			int k2;
			if (k1 == k)
				k2 = k;
			else
			{
				for (int i = k; i < k1 + 1; i++)
					if (cans[i] == 0)
					{
					if (i == 0)
						cans[i] = 1;
					else
					{
						if (ratings[i] <= ratings[i - 1])
							cans[i] = 1;
						else
						cans[i] = cans[i - 1] + 1;
					}
					}
				k1++;
				k2 = k1 ;
			}
			while (k2 + 1 < ratings.size() && ratings[k2] > ratings[k2 + 1])
				k2++;
			if (k1<ratings.size()&&k1>0 && k2 + 1 - k1 >= cans[k1 - 1] && ratings[k1 - 1] > ratings[k1])
				cans[k1 - 1] = k2 + 1 - k1 + 1;
			for (int i = k1; k1<ratings.size()&&i < k2 + 1; i++)
				cans[i] = k2+1 - k1  - (i - k1);
			if (k2 + 1 >= ratings.size())
				break;
			if (ratings[k2] == ratings[k2 + 1])
				k = k2 + 1;
			else
				k = k2;
		}
		int sum = 0;
		for (int i = 0; i < cans.size(); i++)
			sum += cans[i];
		return sum;
	}
};