摘要:保證高的小朋友拿到的糖果更多,我們建立一個分糖果數組。首先,分析邊界條件如果沒有小朋友,或者只有一個小朋友,分別對應沒有糖果,和有一個糖果。排排坐,吃果果。先往右,再往左。右邊高,多一個。總和加上小朋友總數,就是要準備糖果的總數啦。
Problem
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?
ExampleGiven ratings = [1, 2], return 3.
Given ratings = [1, 1, 1], return 3.
Given ratings = [1, 2, 2], return 4. ([1,2,1]).
Note保證rating高的小朋友拿到的糖果更多,我們建立一個分糖果數組A。小朋友不吃糖果,考試拿A也是可以的。
首先,分析邊界條件:如果沒有小朋友,或者只有一個小朋友,分別對應沒有糖果,和有一個糖果。
然后我們用兩個for循環來更新分糖果數組A。
排排坐,吃果果。先往右,再往左。右邊高,多一個。左邊高,吃得多。
這樣,糖果數組可能就變成了類似于[0, 1, 0, 1, 2, 3, 0, 2, 1, 0],小朋友們一定看出來了,這個不是真正的糖果數,而是根據考試級別ratings給高分小朋友的bonus。
好啦,每個小朋友至少要有一個糖果呢。
bonus總和加上小朋友總數n,就是要準備糖果的總數啦。
</>復制代碼
public class Solution {
public int candy(int[] ratings) {
int n = ratings.length;
if (n < 2) return n;
int[] A = new int[n];
for (int i = 1; i < n; i++) {
if (ratings[i] > ratings[i-1]) A[i] = A[i-1]+1;
}
for (int i = n-2; i >= 0; i--) {
if (ratings[i] > ratings[i+1]) A[i] = Math.max(A[i], A[i+1]+1);
}
int res = n;
for (int a: A) res += a;
return res;
}
}
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