Provide all my solutions and explanations in Chinese for all the Leetcode coding problems. Same as this: LeetCode All in One 题目讲解汇总(持续更新中...) Click below image to watch YouTube Video Note: All explanations are written in Github Issues, please do not create any new issue or ...
This repository contains the solutions and explanations to the algorithm problems on LeetCode. Only medium or above are included. All are written in C++/Python and implemented by myself. The problems attempted multiple times are labelled with hyperlinks. Topics algorithms leetcode cpp Resources Re...
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// Solution 2: This is a clever solution. (I'm not saying it's a short solution.) Took me more than an hour to figure out. First of all, I believed there's a better solution to this problem than an O(N ^ 2) brute-force one. Because it's about "order", and you know there...
After, we can build a divide and conquer. Suppose we are looking at the interval S[i]...S[j] and it is supposed to represent L, R. Let's see whereM = (L+R)//2is supposed to start assuming the representation of [L, M-1] is untouched. We need to know how many digits are...
count(sum[k-1]-sum[j])) return true; } } return false; } }; 572 ms 11 MB 参考大佬的题解剪枝 作者:palading 链接:https://leetcode-cn.com/problems/split-array-with-equal-sum/solution/c-qian-zhui-he-jian-zhi-by-palading/ 来源:力扣(LeetCode) 著作权归作者所有。商业转载请联系作者...
[i + rowSize - 1][k] = num++;53}5455for(intk = i + rowSize - 2; k > i; k--) {//both the start and end need to be i, j, and also care about length56matrix[k][j] = num++;57}5859this.generateMatrixHelper(i + 1, j + 1, rowSize - 2, colSize - 2, matrix, ...
解法1:dp[i][j]表示1~(i+1)排列中逆序数为j的种类数,动规搞起。 // Solution 1: Let dp[i][j] be the number ofpermutationswith inverse number j, you know what to do. 代码1 //Code 1 630 Course Schedule III // #630 课表3
; for (int i = 0; i < n; ++i) // 沿着副对角线反转 for (int j = 0; j <...
0 <= mat[i][j] <= 10000 0 <= threshold <= 10^5 Problemlink Video Tutorial You can find the detailed video tutorial here Youtube B站 Thought Process There is an absolute brute force way is to calculate the sum of each square in every iteration. There are previous similar problems that...