LeetCode-135. Candy (JAVA)根据等级分糖果

135. Candy(贪心)

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。

时间复杂度：O(n)

空间复杂度：O(n)

实现：

1、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；

2、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；

3、对A求和即为最少需要的糖果。

public int candy(int[] ratings) {int A[] = new int[ratings.length];// 每人至少一个糖果Arrays.fill(A, 1);for (int i = 1; i < ratings.length; i++)if (ratings[i] > ratings[i - 1])A[i] = A[i - 1] + 1;int sum = A[ratings.length - 1];for (int i = ratings.length - 2; i >= 0; i--) {if (ratings[i] > ratings[i + 1] && A[i] <= A[i + 1])// 要比邻居多,并求与邻居相比的取较大值A[i] = A[i + 1] + 1;sum += A[i];}return sum;}

// 分别从两边遍历选取较大值，计算结果public int candy(int[] ratings) {int L[] = new int[ratings.length];int R[] = new int[ratings.length];// 每人至少一个糖果Arrays.fill(L, 1);Arrays.fill(R, 1);for (int i = 1; i < ratings.length; i++)if (ratings[i] > ratings[i - 1])L[i] = L[i - 1] + 1;for (int i = ratings.length - 2; i >= 0; i--) {if (ratings[i] > ratings[i + 1])R[i] = R[i + 1] + 1;}int sum = 0;for (int i = 0; i < ratings.length; i++)sum += Math.max(L[i], R[i]);return sum;}