Candy

https://leetcode.com/problems/candy/
给定一个int数组，每个int代表一个小孩手中的数字，给每个小孩分糖果，保证相邻的小孩中，手中数字大的小孩拿到的糖果大于手中数字小的小孩，请问最终最少需要多少糖果。(每个小孩至少有一个糖果)

• Each child must have at least one candy.
• Children with a higher rating get more candies than their neighbors.
• 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.

解题思路

单独一个数组dp记录每个小孩的糖果数量

• 初始化每个小孩的糖果数量为1
• 正向遍历，保证邻居下一个小孩如果数字大，糖果一定高
• 反向遍历，保证上一个小孩如果数字大，糖果一定高
• 反向遍历的时候有个特例的情况,需要保证i-1的大小不会完全收到i的限制dp[i - 1] = Math.max(dp[i - 1], dp[i] + 1);,用这个例子可以看看区别：input = [1,3,4,5,2]

上代码

public int candy(int[] ratings) {
        int len = ratings.length;
        int[] dp = new int[len];
        Arrays.fill(dp, 1);
        for (int i = 0; i < len - 1; i++) {
            if (ratings[i + 1] > ratings[i]) {
                dp[i + 1] = dp[i] + 1;
            }
        }

        for (int i = len - 1; i > 0; i--) {
            if (ratings[i - 1] > ratings[i]) {
                dp[i - 1] = Math.max(dp[i - 1], dp[i] + 1);
//                dp[i - 1] = dp[i] + 1; //这个不对，input 用[1,3,4,5,2] 可以看出端倪
            }
        }

        int res = 0;
        for (int i = 0; i < len; i++) {
            res += dp[i];
        }
        return res;
    }