题意是给一个长度为n的字符串，然后要缩短这个字符串，字符串中UR或者RU的长度可以缩为1，求最短长度。
题目要求 Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal
问题： Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal
result -= mat[mid][mid] return result Reference https://leetcode.com/problems/matrix-diagonal-sum
else: result += temp return result Reference https://leetcode.com/problems/diagonal-traverse
题目 每次把空列换到最后一列，把非空行换到最下一行。 #include<cstdio> #include<algorithm> #define N 1005 u...
diagonals.append([]) diagonals[i+j].append(nums[i][j]) result = [] for diagonal...in diagonals: result += diagonal[::-1] return result Version 3 class Solution:...queue.append((i, j+1)) return result Reference https://leetcode.com/problems/diagonal-traverse-ii
Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order
Below the Diagonal You are given a square matrix consisting of n rows and n columns....In that special form all the ones must be in the cells that lie below the main diagonal....matrix, which is located on the intersection of the i-th row and of the j-th column, lies below the main diagonal
Diagonal的激动剂抗体能激活HHT患者基因受损的TGF-β超家族受体复合物。在HHT的临床前模型中，Diagonal的激动剂抗体可以预防和逆转病理性血管畸形的形成。...DIAGONAL平台解决了这一难题，它结合了专有的计算和实验技术，使Diagonal的科学家能够以前所未有的速度高保真地筛选数十亿种组合。...Diagonal的激动剂抗体能纠正信号传导，恢复血管壁的平衡。...关于Diagonal Therapeutics Diagonal Therapeutics是一家生物技术公司，开创了发现和开发激动剂抗体的新方法。...利用DIAGONAL平台发现的Diagonal新兴产品线有望通过解决疾病的根本原因为患者提供改变生命的疗法。
Diagonal连接方式 3. FULL_CONTACT连接方式 4. 8_WAY连接方式 ---- 1....Diagonal连接方式 Diagonal：英文含义是斜的，特点是焊盘到铜皮的Cline最多4条，且Cline之间互相不垂直。...下图Smd pins设置为Diagonal时，Minimum connects 设置为4，Maximum connects 设置为4后的效果。 3....FULL_CONTACT连接方式 似乎也没什么好解释的... 4. 8_WAY连接方式 8_WAY就是最多8条cline连接焊盘和铜皮，实际是Orthogonal和Diagonal组合成的一种连接方式
grid, i, j, k): return k def check(self, grid, x, y, k): main_diagonal...= 0 secondary_diagonal = 0 for i in range(k): main_diagonal += grid[x+i]...[y+i] secondary_diagonal += grid[x+i][y+k-i-1] if main_diagonal !...= secondary_diagonal: return False for i in range(k): row = sum(grid...= main_diagonal: return False column = sum([grid[x+j][y+i] for j in range
; DiagonalMatrix 结构体定义了对角矩阵的结构，包括矩阵的维度 size 和存储对角元素的数组 diagonal。...= col) { printf("Error: Only diagonal elements can be set....\n"); } else { matrix->diagonal[row] = value; } } setElement 函数用于设置对角矩阵中指定位置的元素值，接受一个指向...= col) { printf("Error: Only diagonal elements can be set....\n"); } else { matrix->diagonal[row] = value; } } // 获取对角矩阵中指定位置的元素值 int getElement(
reuse_diagonal_) { if (diagonal_.rows() !...= num_parameters) { diagonal_.resize(num_parameters, 1); } jacobian->SquaredColumnNorm(diagonal_.data...()); for (int i = 0; i < num_parameters; ++i) { diagonal_[i] = std::min(std::max(diagonal_[i], min_diagonal..._), max_diagonal_); } } // D = diag{J'*J}/radius lm_diagonal_ = (diagonal_ / radius_).array().sqrt();...* step_quality - 1.0, 3)); radius_ = std::min(max_radius_, radius_); decrease_factor_ = 2.0; reuse_diagonal
()代码 private boolean isWin () { int total = lastPlayer * SIZE; char diagonal1...= '\0'; char diagonal2 = '\0'; for (int index = 0; index < SIZE; index++) {...+= board[index][index]; diagonal2 += board[index][SIZE - index - 1]; } /...= '\0'; char diagonal2 = '\0'; for (int index = 0; index diagonal1 == total || diagonal2 == total; } public char nextPlayer () {
A fairy came and told her how to solve this puzzle lock: “When the sum of main diagonal and anti-diagonal...Here, main diagonal is the diagonal runs from the top left corner to the bottom right corner, and anti-diagonal
= row - i; // 如果该斜线上已经有了皇后则进行下一个 for 循环 if (diagonals1.contains(diagonal1...= row + i; // 如果该斜线上已经有了皇后则进行下一个 for 循环 if (diagonals2.contains(diagonal2...i; // 记录入受影响的列和两个斜线 columns.add(i); diagonals1.add(diagonal1...); diagonals2.add(diagonal2); // 递归主体 backtrack(solutions...); diagonals2.remove(diagonal2); } } } // 生成结果棋盘的方法
grid[i][j]] == 1: return False temp[grid[i][j]] = 1 main_diagonal...= 0 secondary_diagonal = 0 for i in range(k): main_diagonal += grid[x+i]...[y+i] secondary_diagonal += grid[x+i][y+k-i-1] if main_diagonal !...= secondary_diagonal: return False for i in range(k): if main_diagonal...= sum(grid[x+i][y:y+k]): return False if main_diagonal !
The argument diagonal controls which diagonal to consider: If diagonal = 0, it is the main diagonal....If diagonal > 0, it is above the main diagonal....The argument diagonal controls which diagonal to consider....If diagonal = 0, all elements on and below the main diagonal are retained....The argument diagonal controls which diagonal to consider.
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