标签：style blog http io ar color sp for on
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:
What is the minimum candies you must give?
Solution:
基本思路就是进行两次扫描，一次从左往右，一次从右往左。第一次扫描的时候维护对于每一个小孩左边所需要最少的糖果数量，存入数组对应元素中，第二次扫描的时候维护右边所需的最少糖果数，并且比较将左边和右边大的糖果数量存入结果数组对应元素中。这样两遍扫描之后就可以得到每一个所需要的最最少糖果量，从而累加得出结果。方法只需要两次扫描，所以时间复杂度是O(2*n)=O(n)。空间上需要一个长度为n的数组，复杂度是O(n)。
1 public class Solution { 2 public int candy(int[] ratings) { 3 if(ratings.length==0) 4 return 0; 5 int N=ratings.length; 6 int[] candies=new int[N]; 7 candies[0]=1; 8 for(int i=1;i<N;++i){ 9 if(ratings[i]>ratings[i-1]){ 10 candies[i]=candies[i-1]+1; 11 }else{ 12 candies[i]=1; 13 } 14 } 15 for(int i=N-2;i>=0;--i){ 16 if(ratings[i]>ratings[i+1]&&candies[i]<=candies[i+1]) 17 candies[i]=candies[i+1]+1; 18 } 19 int result=0; 20 for(int iCandy:candies){ 21 result+=iCandy; 22 } 23 return result; 24 } 25 }
标签：style blog http io ar color sp for on
原文地址：http://www.cnblogs.com/Phoebe815/p/4118683.html
