2018-09-27 23:33:49 问题描述: 问题求解: 方法一.DP(MLE) 动态规划的想法应该是比较容易想到的解法了,因为非常的直观,但是本题的数据规模还是比较大的,如果直接使用动态规划,即使不MLE,也是肯定会在大规模的数据量上TLE的. public int sumSubarrayMins(int[] A) { int res = 0; int mod = (int)Math.pow(10, 9) + 7; int[][] dp = new int[A.length][A.len…

Given an array of integers A, find the sum of min(B), where B ranges over every (contiguous) subarray of A. Since the answer may be large, return the answer modulo 10^9 + 7. Example 1: Input: [3,1,2,4] Output: 17 Explanation: Subarrays are [3], [1],…

Given an array of integers A, find the sum of min(B), where B ranges over every (contiguous) subarray of A. Since the answer may be large, return the answer modulo 10^9 + 7. Example 1: Input: [3,1,2,4] Output: 17 Explanation: Subarrays are [3], [1],…

643. 子数组最大平均数 I 643. Maximum Average Subarray I 题目描述 给定 n 个整数,找出平均数最大且长度为 k 的连续子数组,并输出该最大平均数. LeetCode643. Maximum Average Subarray I 示例 1: 输入: [1,12,-5,-6,50,3], k = 4 输出: 12.75 解释: 最大平均数 (12-5-6+50)/4 = 51/4 = 12.75 注意: 1 <= k <= n <= 30,000. 所…

作者: 负雪明烛 id: fuxuemingzhu 公众号:每日算法题 目录 题目描述 题目大意 解题方法 方法一:preSum 方法二:滑动窗口 刷题心得 日期 题目地址:https://leetcode.com/problems/maximum-average-subarray-i/description/ 题目描述 Given an array consisting of n integers, find the contiguous subarray of given length k…

Given an array consisting of n integers, find the contiguous subarray of given length k that has the maximum average value. And you need to output the maximum average value. Example 1: Input: [1,12,-5,-6,50,3], k = 4 Output: 12.75 Explanation: Maximu…

In an array A of 0s and 1s, how many non-empty subarrays have sum S? Example 1: Input: A = [1,0,1,0,1], S = 2 Output: 4 Explanation: The 4 subarrays are bolded below: [1,0,1,0,1] [1,0,1,0,1] [1,0,1,0,1] [1,0,1,0,1] Note: A.length <= 30000 0 <= S <…

Given an array of integers A, find the sum of min(B), where B ranges over every (contiguous) subarray of A. Since the answer may be large, return the answer modulo 10^9 + 7. Example 1: Input: [3,1,2,4] Output: 17 Explanation: Subarrays are [3], [1],…

题目如下: 解题思路:我的想法对于数组中任意一个元素,找出其左右两边最近的小于自己的元素.例如[1,3,2,4,5,1],元素2左边比自己小的元素是1,那么大于自己的区间就是[3],右边的区间就是[4,5].那么对于元素2来说,和左边区间合并组成[2,3]以及和右边区间合并组成[2,4,5],这两段区间包括2在内的所有subarray的最小值都是2,其分别可以组成的subarry的个数是 len([3])和len([4,5]),左右区间合并在一起可以组成的subarray的个数是len([3])…

思路: 对于每个数字A[i],使用单调栈找到A[i]作为最小值的所有区间数量,相乘并累加结果.时间复杂度O(n). 实现: class Solution { public: int sumSubarrayMins(vector<int>& A) { ; stack<int> st; int n = A.size(); ; ; i < n; i++) { while (!st.empty() && A[i] < A[st.top()]) { int…