题目描述:
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?
N个孩子站成一排,给每个人设定一个权重(已知)。按照如下的规则分配糖果: (1)每个孩子至少分得一颗糖果 (2)权重较高的孩子,会比他的邻居获得更多的糖果。
问:总共最少需要多少颗糖果?请分析算法思路,以及算法的时间,空间复杂度是多少。
假设每个孩子分到的糖果数组为A[N],初始化为{1},因为每个人至少分到一颗糖。
思路解析:
时间复杂度:O(n) ;空间复杂度:O(n)
- 与前面的邻居比较,前向遍历权重数组ratings,如果ratings[i]>ratings[i-1],则A[i]=A[i-1]+1;
- 与后面的邻居比较,后向遍历权重数组ratings,如果ratings[i]>ratings[i+1]且A[i]<A[i+1]+1,则更新A,A[i]=A[i+1]+1;
- 对A求和即为最少需要的糖果。
代码:
public class Solution { public int candy(int[] ratings) { if(ratings==null||ratings.length==0){//第一步:判空 return 0; } int len = ratings.length; int[] candy = new int[len]; candy[0]=1; //从左至右,右边的比左边的大,就可以把右边的糖果数量设置为左边的加1 for(int i=1;i<len;i++){ if(ratings[i]>ratings[i-1]){ candy[i]=candy[i-1]+1; }else{ candy[i]=1; } } //从右边到左边,左边比右边的大,并且原来左边的糖果数量小于等于右边的,就可以把左边的糖果数量设置为右边的加1; for(int i=len-2;i>=0;i--){ if((ratings[i]>ratings[i+1])&& (candy[i]<=candy[i+1])){ candy[i]=candy[i+1]+1; } } int res=0; for(int i=0;i<len;i++){ res+=candy[i]; } return res; } }