(n +1) /2: (m +1) /2;8intp = m, q =n;9for(inti =0; i < c; ++i, p -=2, q -=2) {10for(intcol = i; col < i + q; ++col)11res.push_back(matrix[i][col]);12for(introw = i +1; row < i + p; ++row)13res.push_back(matrix[row][i + q -1]);14if(...
(n +1) /2: (m +1) /2;8intp = m, q =n;9for(inti =0; i < c; ++i, p -=2, q -=2) {10for(intcol = i; col < i + q; ++col)11res.push_back(matrix[i][col]);12for(introw = i +1; row < i + p; ++row)13res.push_back(matrix[row][i + q -1]);14if(...
di= (di + 1) % 4r, c= r + dr[di], c +dc[di]returnans 按层模拟 思路: 一层一层遍历。 代码: classSolution(object):defspiralOrder(self, matrix):defspiral_coords(r1, c1, r2, c2):forcinrange(c1, c2 + 1):yieldr1, cforrinrange(r1 + 1, r2 + 1):yieldr, c2ifr1 < r2and...
problem 54. Spiral Matrix solution #1: traverse from left to right, and then from down to up; code class Solution { public: vector<int> spiralOrder(vector<vector<int>>& matrix) { vector<int> res; if(matrix.empty()) return res; ...
= null){ Character c = queue.peek(); if(map.get(c) == 1) return c; //只有重复的才会出列，不重复会一直保存 else queue.poll(); } return '#'; }} 剑指 21.栈的压入、弹出序列 输入两个整数序列，第一个序列表示栈的压入顺序，请判断第二个序列是否可能为该栈的弹出...
在CV界，传统卷积已经彻底输给Transformer了吗？连续
这里要注意,string res; atoi(res.c_str()); atoi是将字符串数组转整型,若是string,res.c_str() if (s.find("sjjd") != string::npos)查找操作,找到,则返回子串第一个字符的位置,没找到,则返回string::npos。 s = s.substr(0,4)截取操作,从下标0开始截取连续4个字符 ...
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push_back(matrix[r1][c]);//像右遍历行 for (int r = r1 + 1; r <= r2; r++) res.push_back(matrix[r][c2]);//向下遍历列 if (r1 < r2 && c1 < c2) { for (int c = c2 - 1; c > c1; c--) res.push_back(matrix[r2][c]);//向左遍历行 for (int r = r2; r > ...
(va.second + vb.second,min(va.second,vb.second) + 1)}); }else if(op.top() == '|'){ states.push({min(va.first + vb.first,min(va.first,vb.first) + 1),min(va.second,vb.second)}); } op.pop(); } }else{ op.push(c); } } return max(states.top().first,states.top(...