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?
Example 1:
Input: [1,0,2]
Output: 5
Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
Example 2:
Input: [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
The third child gets 1 candy because it satisfies the above two conditions.

實現題（貪婪法）
1.給所有孩子一顆糖。
2.從左到右遍歷，把當前分數和左邊比，如果高了，比左邊的糖多一顆。
3.從右到左遍歷，把當前分數和右邊比，如果高了，比右邊的糖多一顆。
4.再掃一遍計數。

細節：
1.第三步的時候，如果要和右邊比，一定要從右到左掃，不能從左到右掃。如果從左到右，因為你右邊的對象還沒有跟右邊的右邊的對象對比過，它的值都還不是確定的，你怎麽能直接拿過來計算結果呢，錯誤。比如分數321，在和右邊數對比的時候，從左到右掃就會是111->111->221錯誤，從右到左掃會是111->111->321正確
2.給所有孩子一顆糖有優雅點的寫法: Arrays.fill(array, 1);

實現：

class Solution {
    public int candy(int[] ratings) {
        if (ratings == null || ratings.length == 0) {
            return 0;
        }
        int[] candies = new int[ratings.length];
        
        // P2: 一個優雅的填充寫法。
        Arrays.fill(candies, 1);
        // for (int i = 0; i < candies.length; i++) {
        
 
//     candies[i] = 1;
        // }

        for (int i = 1; i < candies.length; i++) {
            if (ratings[i] > ratings[i - 1] && candies[i] <= candies[i - 1]) {
                candies[i] = candies[i- 1] + 1;
            }
        }
        
        // P1: 和右邊的數字對比的時候，要從右到左遍歷，否則會出錯，比如321的孩子，在和右邊數對比的時候，從左到右掃就會是111->111->221，從右到左掃會是111->111->321正確
        for (int i = candies.length - 2; i >= 0; i--) {
            if (ratings[i] > ratings[i + 1] && candies[i] <= candies[i + 1]) {
                candies[i] = candies[i + 1] + 1;
            }
        }
        
        int ans = 0;
        for (int i = 0; i < candies.length; i++) {
            ans += candies[i];
        }
        return ans;
    }
}

leetcode135 - Candy - hard