LeetCode Candy

最新推荐文章于 2023-02-28 14:31:32 发布

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本文介绍了一个算法问题——如何以最少的糖果数量满足每个孩子的糖果需求，确保评分高的孩子比邻居获得更多的糖果。通过一个时间复杂度为O(n)且空间复杂度为O(1)的C++实现方案来解决该问题。

题目：

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?

解法及分析参考： http://www.cnblogs.com/felixfang/p/3620086.html

class Solution {
public:
	//时间复杂度为O(n)，空间复杂度为O(1)
	int candy(vector<int> &ratings) {
		int n = ratings.size();
		if (n == 0)
			return 0;
		int res = 0, pre = 1, cur;
		//mark记录严格递减序列的开始
		int mark = pre;
		//len表示递减序列的长度
		int len = 0;
		res++;
		for (int i = 1; i < n; i++) {
			if (ratings[i] < ratings[i - 1]) {
				len++;
				if (len >= mark)
					res++;
				res += len;
				pre = 1;				
			}
			else {
				if (ratings[i] > ratings[i - 1])
					cur = pre + 1;
				else
					cur = 1;
				res += cur;
				len = 0;
				pre = cur;
				mark = pre;
			}
		}
		return res;
	}
};