Leetcode:Candy 每个数都比相邻数大

原题戳我

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?

class Solution {
public:
    int candy(vector<int> &ratings) {
        std::vector<int> num(ratings.size(), 1);

        int inc = 2;
        for (int i = 1; i < ratings.size(); ++i) {
            if (ratings.at(i) > ratings.at(i - 1)) {
                num.at(i) = std::max(inc++, num.at(i));
            } else {
                inc = 2;
            }
        }
        inc = 2;
        for (int i = ratings.size() - 2; i >= 0; --i) {
            if (ratings.at(i) > ratings.at(i + 1)) {
                num.at(i) = std::max(inc++, num.at(i));
            } else {
                inc = 2;
            }
        }
        int res = std::accumulate(num.begin(), num.end(), 0);
        return res;
    }
};

step1. 首先将辅助数组每个元素初始化为1，因为每个元素都至少为1

step2. 从左向右扫描，如果某个数比其左相邻数大，则num数组对应位置的这个数从2开始递增赋值 (注意相等是不需要增加的)，否则递增量初始为2

遍历完之后，可以满足：如果 原数组中某个数大于其左相邻数，那么对应辅助数组中也一定是大于其左相邻数

如果 ratings[i] >= ratings[i+1] 此时ratings[i+1]一定为1

step3. 从右向左遍历，类似step2

其实step2中 num.at(i)一定是等于inc++的，因为num.at(i)未改变之前都是初始为1的

为什么step3中一定要加上 inc比num.at(i)的比较呢？ 因为此时 必须满足step2之后数组的大小条件

原数组为：1 --> 2 --> 5 --> 4 --> 3 --> 2 --> 1

step2之后 num数组为：1 --> 2 --> 3 --> 1 --> 1 --> 1 --> 1

step3从右向左遍历到idx=2时 : 1 --> 2 --> （3 or 5）--> 4 --> 3 --> 2 --> 1

idx=2位置处，必须满足它比左相邻数大，比右相邻数大，所以才有了max操作