解答不需要数组。 classSolution {public:intmaxSubArray(vector<int>&nums) {//明显的动态规划,因为是连续的,所以这个子序列:1.取前面和后面的元素则和变小,//比如,如果4,左边的dp(取该元素)是依次[-2,1,-2,4,3,5,6,1,5][]if(nums.empty())return0; vector<int>vec; vec.push_back(nums[0])...
写在前面 这题被标记为hard,第一眼看到,确实很容易想到是dp,但思路之后就陷入混乱,原因就是不知道dp的状态转移方程,以及dp过程的开始和结束,本题事实上是两道题目的组合,分别是 Max Sum of Rectangle in a matrix,这道题的题解可以看这个视频(需要翻墙) 以及这道题目:max subarray sum no more than k 本...
这个问题是这样的(https://leetcode.com/problems/maximum-subarray/): Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. 就是给一组整数,找到一组连续的子序列,让子序列的和最大,并且返回最大和。 比如说: 输入...
Maximum Subarray Range Sum Query 2D - Immutable Maximum Size Subarray Sum Equals k 参考资料: https://leetcode.com/problems/max-sum-of-rectangle-no-larger-than-k/ https://leetcode.com/problems/max-sum-of-rectangle-no-larger-than-k/discuss/83618/2-Accepted-Java-Solution https://leetcode.com...
Leetcode之Maximum Subarray 问题 问题描述: Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array [-2,1,-3,4,-1,2,1,-5,4], the contiguous subarray [4,-1,2,1] ......
0152-Maximum-Product-Subarray 0153-Find-Minimum-in-Rotated-Sorted-Array 0155-Min-Stack 0159-Longest-Substring-with-At-Most-Two-Distinct-Characters 0160-Intersection-of-Two-Linked-Lists 0161-One-Edit-Distance 0167-Two-Sum-II-Input-array-is-sorted 0169-Majority-Element 01...
First thing to note is that sum of subarray(i,j]is just the sum of the firstj elements less the sum of the firsti elements. Store these cumulative sums in the array cum. Then the problem reduces to findingi,j such thati<j andcum[j]−cum[i] is as close tokbut lower than it....
count(distinct(CountryCode)) 232 是不是看到數字瞬間變小許多了呢?這是因為 DISTINCT() 函式不會計算重複的值。 SUM 我覺得 SUM 得跟 COUNT 放在一起講,我當初剛寫的時候很常把這兩個指令搞混: SUM() 是全部資料的『值』加總 COUNT() 是計算查找欄位總共有多少『筆』資料 ...
// you cannot add this in this subarray, make new one // say you add this num in new subarray, then sum = num sum = num; pieces++; } else { sum += num; } }if (pieces > m) { start = mid + 1; } // else if (pieces <= m) { end = mid }; ...
leetcode 695 岛屿的最大面积 max-area-of-island【ct】,思路:遍历每一个节点,如果点是1,就开始计算当前岛屿的数量,传入一个ref到dfs函数中用于计算,dfs函数首先设置退出条件,其次将该点的值设置为0,增加ref的值,向四个方向继续dfs