1 题目
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?
2 思路
按照网上的思路，每个孩子至少有一个糖果，先从左到右遍历一遍，写出递增的糖果数，再从右到左遍历一遍完成递减的糖果数。这种大小与左右两边数据相关的问题，均可以采用这个思路。另外，有另一种空间复杂度O（1），时间复杂度O（n）的思路，可以参考http://www.cnblogs.com/felixfang/p/3620086.html。
3 代码
public int candy(int[] ratings) { if(ratings == null || ratings.length == 0) { return 0; } int[] candyNums = new int[ratings.length]; candyNums[0] = 1; for(int i = 1; i < ratings.length; i++) { if(ratings[i] > ratings[i-1]) //如果第i个孩子比第i - 1孩子等级高， { candyNums[i] = candyNums[i-1]+1; } else //每人至少有一个糖果 { candyNums[i] = 1; } } for(int i = ratings.length-2; i >= 0; i--) { if(ratings[i] > ratings[i + 1] && candyNums[i] <= candyNums[ i + 1]) //如果第i个孩子比第i + 1孩子等级高并且糖果比i+1糖果少 { candyNums[i] = candyNums[i + 1] + 1; } } int total = 0; for (int i = 0; i < candyNums.length; i++) { total += candyNums[i]; } return total; }
原文：http://www.cnblogs.com/lingtingvfengsheng/p/4304505.html