这个题对我来说真的是相当难的题目了,严格来讲可能不算是个动态规划的题目,但这个题目对类似的划分多个非重叠连续子区间的问题提供了一个很好解决方案 这个题目需要找三个非重叠的连续子区间,通过维护两个数组将第一个和第三个子区间可能的开始pos记录下来,在中间那个子区间开始的pos遍历时限制其边界范围,根据长度能恰到好处的将三个子区间划分成非重叠的,还使用了集合相减代替累加这样比较简单好用的技巧 下面提供代码: class Solution { public int[] maxSumOfThreeSub…
2020-02-18 20:57:58 一.Maximum Subarray 经典的动态规划问题. 问题描述: 问题求解: public int maxSubArray(int[] nums) { int res = nums[0]; int n = nums.length; int[] dp = new int[n]; dp[0] = nums[0]; for (int i = 1; i < n; i++) { if (dp[i - 1] < 0) dp[i] = nums[i]; else…
1146. Maximum Sum Time limit: 1.0 second Memory limit: 64 MB Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this…
Maximum sum Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 40596 Accepted: 12663 Description Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below: Your task is to calculate d(A). Input The input consists o…
In a given array nums of positive integers, find three non-overlapping subarrays with maximum sum. Each subarray will be of size k, and we want to maximize the sum of all 3*k entries. Return the result as a list of indices representing the starting p…
Maximum sum Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 39599 Accepted: 12370 Description Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below: Your task is to calculate d(A). Input The input consists o…
1146. Maximum Sum Time limit: 0.5 secondMemory limit: 64 MB Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this p…
题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problem=44 Maximum Sum Background A problem that is simple to solve in one dimension is often much more difficult to solve in more th…
Maximum Sum 大意:给你一个n*n的矩阵,求最大的子矩阵的和是多少. 思路:最開始我想的是预处理矩阵,遍历子矩阵的端点,发现复杂度是O(n^4).就不知道该怎么办了.问了一下,是压缩矩阵,转换成最大字段和的问题. 压缩行或者列都是能够的. int n, m, x, y, T, t; int Map[1010][1010]; int main() { while(~scanf("%d", &n)) { memset(Map, 0, sizeof(Map)); for(i…
