算法第十三週作業01

Description

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?

Solution

• 首先理解題意：如果該孩子的權值比傍邊孩子的大，那麼他的糖果數比傍邊孩子的要多，對於權值爲[1,2,2]情況，第一個孩子糖果數爲1；第二個孩子權值比第一個孩子權值大，故爲2；第三個孩子糖果樹不比前面孩子（只有前面鄰居）多，故糖果樹爲1
• 分別考慮權值的正向上升和反向上升。上升過程，如果對應糖果數arr[i]比待比較糖果數arr[i-1]大，則arr[i]保持不變，arr[i]=arr[i-1]+1。
• 對於權值序列[1,2,3,6,5,2,3,3,4]，首先輸出化糖果樹序列[0,0,0,0,0,0,0,0,0] （PS：題目中要求每個孩子至少1個糖果，爲避免初始化序列浪費O(n)複雜度，此處不進行手動初始化，有後面統一加上序列長度來彌補），對於正向上升，糖果樹序列爲[0,1,2,3,0,0,1,0,1]，對於反向上升，糖果序列變換爲[0,1,2,3,1,0,1,0,1]

Code

public class Solution {
    public int candy(int[] ratings) {
        int result = 0;
        int[] arr = new int[ratings.length];
        // 正向上升
        for (int i = 1; i < ratings.length; i++) {
            // 相鄰對比
            if(ratings[i] > ratings[i-1]){
                arr[i] = arr[i-1] + 1;
            }
        }
        // 反向上升
        for (int i = ratings.length - 1; i > 0; i--) {
            // 相鄰對比，如果權值遞增時，糖果數也是遞增的，則忽略操作
            if(ratings[i-1] > ratings[i] && arr[i-1] <= arr[i]){
                arr[i-1] = arr[i] + 1;
            }
            // 疊加結果
            result += arr[i];
        }
        return result + arr[0] + ratings.length;
    }
}