【LeetCode】135.Candy(hard)解题报告

【LeetCode】135.Candy(hard)解题报告

tags: greedy

题目地址：https://leetcode.com/problems/candy/description/

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:

  1. Each child must have at least one candy.

  2. Children with a higher rating get more candies than their neighbors.

  What is the minimum candies you must give?

题意：分糖，看怎么分需要的糖最少，满足每个人至少一个糖，数大的人比左右两边糖多，遍历两次，正序一次，倒序一次，倒序时要注意就是如果左面值比右面加一的大，就要取原有值，即用到max函数，比较取最大的。

Solutions:

思路：
1 2 3 4 2 1 1
1 1 1 1 1 1 1
f:   1 2 3 4 1 1 1
b:   1 2 3 4 2 1 1
condy[i] = Math.max(candy[i] , candy[i+1]=1)
代码如下：
class Solution {
public int candy(int[] ratings) {
if(ratings.length==0){
return 0;
}
int[] candy = new int[ratings.length];
Arrays.fill(candy,1);
for(int i=1 ; i<ratings.length ; i++){
if(ratings[i] > ratings[i-1]){
candy[i] = candy[i-1] + 1;
}
}
for(int i=ratings.length-2 ; i>=0 ; i--){
if(ratings[i] > ratings[i+1]){
candy[i] = Math.max(candy[i] , candy[i+1]+1);
}
}
int sum=0;
for(int i=0 ; i<candy.length ;i++){
sum += candy[i];
}
return sum;
}
}

Date：2017年11月6日