leetcode135

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.

实现题（贪婪法）
1.给所有孩子一颗糖。
2.从左到右遍历，把当前分数和左边比，如果高了，比左边的糖多一颗。
3.从右到左遍历，把当前分数和右边比，如果高了，比右边的糖多一颗。
4.再扫一遍计数。

细节：
1.第三步的时候，如果要和右边比，一定要从右到左扫，不能从左到右扫。如果从左到右，因为你右边的对象还没有跟右边的右边的对象对比过，它的值都还不是确定的，你怎么能直接拿过来计算结果呢，错误。比如分数321，在和右边数对比的时候，从左到右扫就会是111->111->221错误，从右到左扫会是111->111->321正确
2.给所有孩子一颗糖有优雅点的写法: Arrays.fill(array, 1);

实现：

class Solution {
    public int candy(int[] ratings) {
        if (ratings == null || ratings.length == 0) {
            return 0;
        }
        int[] candies = new int[ratings.length];
        
        // P2: 一个优雅的填充写法。
        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: 和右边的数字对比的时候，要从右到左遍历，否则会出错，比如321的孩子，在和右边数对比的时候，从左到右扫就会是111->111->221，从右到左扫会是111->111->321正确
        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;
    }
}