[leetcode] 适合打劫银行的日子 -前缀和

题目链接

前缀和思想

用数组$l[]$表示前面有多少个数满足$a[i-1]>=a[i]$
用数组$r[]$表示前面有多少个数满足$a[i]<=a[i+1]$

O(n)遍历之后得到$l[],r[]$
然后O(n)找满足$l[i]>=time\&\&r[i]>=time$的下标存入$vector$并返回
Code:

class Solution {
public:
    vector<int> goodDaysToRobBank(vector<int>& security, int time) {
        int n = security.size();
        vector<int> l(n), r(n);
        for(int i=1;i<n;i++) {
            if(security[i] <= security[i-1]) l[i] = l[i-1] + 1;
            if(security[n-i-1] <= security[n-i]) r[n-i-1] = r[n-i] + 1;
            // cout << l[i] << "   " << r[i] << endl;
        }
        vector<int> ans;
        for(int i=time;i<n-time;i++) {
            if(l[i] >= time && r[i] >= time) ans.push_back(i);
        }
        return ans;
    }
};