Candy [leetcode]

类似于经典题目《滑雪》，给个链接 https://vijos.org/p/1011

使用记忆化搜索，如果已经计算过的直接返回，否则才计算。

查看当前节点左右的节点是否可以向下走，如果可以向下走并且不出界，说明当前节点需要比两边节点更大的一个持有更多的糖，那么两边节点的糖数也是如此计算的，最少的糖数是1，通过最后一句

dp[position]=++maxs

得到。

const int maxn = 100000;
class Solution {
public:
int candy(vector<int> &ratings) {
rsize = ratings.size();
for (int i = 0; i < rsize; ++i) rating[i] = ratings[i];
memset(dp, 0, sizeof(dp));
int sums = 0;
for (int i = 0, n = ratings.size(); i < n; ++i) {
sums += dfs(i);
}

return sums;
}
private:

int rsize;
int dp[maxn], rating[maxn];
int dfs(int position) {
if (dp[position] > 0) return dp[position];
int maxs = 0;
if (position-1 >= 0 && rating[position] > rating[position-1]) maxs = max(maxs, dfs(position-1));
if (position+1 < rsize && rating[position] > rating[position+1]) maxs = max(maxs, dfs(position+1));

return dp[position] = ++maxs;
}
};