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【LeetCode】135. Candy 分發糖果（Hard）（JAVA）

題目地址: https://leetcode.com/problems/candy/

題目描述：

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.
 

題目大意

老師想給孩子們分發糖果，有 N個孩子站成了一條直線，老師會根據每個孩子的表現，預先給他們評分。

你需要按照以下要求，幫助老師給這些孩子分發糖果：

• 每個孩子至少分配到 1 個糖果。
• 相鄰的孩子中，評分高的孩子必須獲得更多的糖果。

那麼這樣下來，老師至少需要準備多少顆糖果呢？

解題方法

1. 遍歷兩次，一次從左到右，一次從右到左，結果儲存到一個一維數組裡
2. 先從左到右，如果當前元素大於前面的元素，前一個的結果 + 1，min[i] = min[i - 1] + 1
3. 然後從右到左，如果當前元素大於前面的元素，前一個的結果 + 1， min[i + 1] + 1，然後和從左到右遍歷的結果比較，取大的值

class Solution {
    public int candy(int[] ratings) {
        if (ratings.length <= 0) return 0;
        int[] dp = new int[ratings.length];
        dp[0] = 1;
        for (int i = 1; i < dp.length; i++) {
            if (ratings[i] > ratings[i - 1]) {
                dp[i] = dp[i - 1] + 1;
            } else {
                dp[i] = 1;
            }
        }
        int res = dp[dp.length - 1];
        for (int i = dp.length - 2; i >= 0; i--) {
            if (ratings[i] > ratings[i + 1]) {
                dp[i] = Math.max(dp[i + 1] + 1, dp[i]);
            }
            res += dp[i];
        }
        return res;
    }
}
 

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