LeetCode:2100.适合打劫银行的日子

模拟：双重循环，从下标为time的开始遍历，到下标length-time-1结束，对每一次循环都对应两个循环，一个往前走，一个往后走，满足条件，则将下标加入到ans数组中。对于上述方法，缺点为有重复判断，类似于BF和KMP的区别。解决办法，用动态规划，算出ni数组(以i结尾的最长下降子数组)和zh数组（以i结尾的最长上升子数组），该数组用于存储以下标结尾的最长子字符串长度，遍历时只需同时满足n[i]>time和zh[i]>time两个条件即可。（ps：吼吼吼吼吼吼吼吼吼吼吼吼吼吼吼吼，我想的方法居然和官方的题解一样，进步了进步了，哈哈哈哈哈哈哈哈哈哈），差别故就是怎么实现了。。。。

C++:

class Solution {
public:
    vector goodDaysToRobBank(vector& security, int time) {
        int n = security.size();
        vector left(n, 0);
        vector right(n, 0);
        for (int i = 1; i < n; ++i) {
            if (security[i] <= security[i-1]) {
                left[i] = left[i-1] + 1;
            }
            if (security[n-i-1] <= security[n-i]) { //判断倒数第i个
                right[n-i-1] = right[n-i] + 1;
            }
        }
        vector ans;
        for (int i = time; i < n-time; ++i) {
            if (left[i]>=time && right[i]>= time) {
                ans.emplace_back(i);
            }
        }
        return ans;
    }
};

python：

class Solution:
    def goodDaysToRobBank(self, security: List[int], time: int) -> List[int]:
        n = len(security);
        left = [0] * n
        right = [0] * n
        for i in range(1, n) :
            if security[i] <= security[i-1]:
                left[i] = left[i-1] + 1
            if security[n-i-1] <= security[n-i]:
                right[n-i-1] = right[n-i] + 1
        ans = []
        for i in range(time, n-time):
            if left[i]>=time and right[i]>= time :
                ans.append(i)
        return ans