[leetcode] 135. Candy

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最新推荐文章于 2024-08-24 19:27:15 发布

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分类专栏： C++ leetcode leetcode题解

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本文探讨了分糖果问题的高效解决方案，旨在确保每个孩子的糖果数满足特定条件：每个孩子至少获得一个糖果，评分较高的孩子得到的糖果比邻居多。通过双向遍历和动态调整策略，文章详细介绍了如何最小化总糖果数。

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Description

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?
Example 1:

Input: [1,0,2]
Output: 5
Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.

Example 2:

Input: [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
             The third child gets 1 candy because it satisfies the above two conditions.

分析

题目的意思是：现在给一群孩子分蛋糕，每个孩子有一个值，现在给孩子分蛋糕，要求每个孩子至少一个蛋糕，得分高的孩子得到的蛋糕必须比相邻的孩子高。

先定义一个技术vector，初始值都为1，然后从左到右，如果遇见当前的rating比前一个相邻的大，就nums[i]=nums[i-1]+1;
然后从右向左，如果遇见当前的rating比后面一个小，且nums[i]<=nums[i+1].求完后遍历求和就是最终的结果。
这种题目首先会把人带入动态规划（恐惧）的误区，感觉是不看答案做不出来，实现后发现并不是，我这里虚心学习一下。

代码

class Solution {
public:
    int candy(vector<int>& ratings) {
        int n=ratings.size();
        vector<int> nums(n,1);
        for(int i=0;i<n-1;i++){
            if(ratings[i+1]>ratings[i]){
                nums[i+1]=nums[i]+1;
            }
        }
        for(int i=n-1;i>0;i--){
            if(ratings[i-1]>ratings[i]){
                nums[i-1]=max(nums[i-1],nums[i]+1);
            }
        }
        int res=0;
        for(int i=0;i<n;i++){
            res+=nums[i];
        }
        return res;
        
    }
};