LeetCode 135. Candy--贪心算法

题目链接

135. 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?

解:

这个题难度是hard,其实比较简单,开一个candy[],从左到右,从右到左遍历一次就搞定了,第一次遍历时,如果后一个的权值大于前一个,则candy数也要大一个,反之,candy数等于1。

第一次遍历后,candy[]中还存在一种违反规则的情况:ratings[i] < ratings[i-1] && candy[i-1] <= candy[i],反向遍历时依次修改。

代码:

class Solution {
public:
    int candy(vector<int>& ratings) {
        int size = ratings.size();
        int *candy = new int[size];
        candy[0] = 1;
        for (int i = 1; i < size; i++) {
            if (ratings[i] > ratings[i-1]) candy[i] = candy[i-1] + 1;
            else candy[i] = 1;
        }
        for (int i = size-1; i > 0; i--) {
            if (ratings[i] < ratings[i-1] && candy[i-1] <= candy[i])
                candy[i-1] = candy[i] + 1;
        }
        int count = 0;
        for (int i = 0; i < size; i++) count += candy[i];
        return count;
    }
};

    原文作者:贪心算法
    原文地址: https://blog.csdn.net/monkey_rose/article/details/78133947
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