Candy_Week7

Candy_Week7

题目：(Candy) ←链接戳这里

题目说明：
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?

难度: Hard

解题思路：
题目意为，若干个孩子站成一排，每个孩子都有rating，比较相邻的孩子，rating高的要比rating低的(相邻的)孩子糖果多。
且每个孩子至少有一颗糖，问最少共需要多少颗糖
思路为：
建立candy数组，里面存放每个孩子的糖果数量
顺序遍历rating数组，当右边的孩子rating高的时候，确保他比左边孩子多一个糖，不然(小于或等于)则设置糖果数为1
逆序遍历rating数组，确认是否存在左边孩子rating比右边孩子高的情况，若存在且糖果数确实少，则为其增加糖果，直到比右边孩子高
两次遍历数组即可得出结果
用了O(n)的时间和O(n)的空间

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    int candy(vector<int>& ratings) {
        int total = 0;
        int length = ratings.size();
        //标记每个孩子手中的糖果数量
        vector<int> candy(length);
        candy[0] = 1;

        //顺序遍历数组，确保当左边孩子比右边孩子rating高时，糖果多1
        for (int i = 1;i < length; i++) {
            if (ratings[i] > ratings[i-1]) {
                candy[i] = candy[i-1] + 1;
            } else {
                candy[i] = 1;
            }
        } 

        //逆序遍历数组，确认是否存在右边孩子比左边孩子rating高情况，若存在，则确保其糖果比左边孩子多1
        for (int i = length-2; i >= 0; i--) {
            if (ratings[i] > ratings[i+1] && candy[i] < (candy[i+1]+1)) {
                candy[i] = candy[i+1] + 1;
            }
            total += candy[i];
        }

        total += candy[length-1];
        return total;
    }
};