算法第十三周作业01

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?

Solution

• 首先理解题意：如果该孩子的权值比傍边孩子的大，那么他的糖果数比傍边孩子的要多，对于权值为[1,2,2]情况，第一个孩子糖果数为1；第二个孩子权值比第一个孩子权值大，故为2；第三个孩子糖果树不比前面孩子（只有前面邻居）多，故糖果树为1
• 分别考虑权值的正向上升和反向上升。上升过程，如果对应糖果数arr[i]比待比较糖果数arr[i-1]大，则arr[i]保持不变，arr[i]=arr[i-1]+1。
• 对于权值序列[1,2,3,6,5,2,3,3,4]，首先输出化糖果树序列[0,0,0,0,0,0,0,0,0] （PS：题目中要求每个孩子至少1个糖果，为避免初始化序列浪费O(n)复杂度，此处不进行手动初始化，有后面统一加上序列长度来弥补），对于正向上升，糖果树序列为[0,1,2,3,0,0,1,0,1]，对于反向上升，糖果序列变换为[0,1,2,3,1,0,1,0,1]

Code

public class Solution {    public int candy(int[] ratings) {        int result = 0;        int[] arr = new int[ratings.length];        // 正向上升        for (int i = 1; i < ratings.length; i++) {            // 相邻对比            if(ratings[i] > ratings[i-1]){                arr[i] = arr[i-1] + 1;            }        }        // 反向上升        for (int i = ratings.length - 1; i > 0; i--) {            // 相邻对比，如果权值递增时，糖果数也是递增的，则忽略操作            if(ratings[i-1] > ratings[i] && arr[i-1] <= arr[i]){                arr[i-1] = arr[i] + 1;            }            // 叠加结果            result += arr[i];        }        return result + arr[0] + ratings.length;    }}