算法题之Leetcode分糖果
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:
- 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?
思路:
首先初始化每个人一个糖果,然后这个算法需要遍历两遍,第一遍从左向右遍历,如果右边的小盆友的等级高,等加一个糖果,这样保证了一个方向上高等级的糖果多。然后再从右向左遍历一遍,如果相邻两个左边的等级高,而左边的糖果又少的话,则左边糖果数为右边糖果数加一。最后再把所有小盆友的糖果数都加起来返回即可。
代码:
public class Solution { public int candy(int[] ratings) { int[] nums = new int[ratings.length]; for (int i=0; i<ratings.length; i++) { nums[i] = 1; } for (int i=0; i<ratings.length - 1; i++) { if (ratings[i+1] > ratings[i]) { nums[i+1] = nums[i] + 1; } } for (int i=ratings.length - 1; i>0; i--) { if (ratings[i-1] > ratings[i]) { nums[i-1] = Math.max(nums[i] + 1, nums[i - 1]); } } int res = 0; for (int i=0; i<ratings.length; i++) { res += nums[i]; } return res; } }