54. Spiral Matrix - cocoder39/coco39_LC GitHub Wiki

Writing clean code under time pressure is challenging

it is easy to miss the 4 break conditions within the while loop, which caused bug eg suppose up is already > down after executing 1st for loop, and right is still >= left then the 3rd for loop will be executed

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        res = []
        up, down, left, right = 0, len(matrix)-1, 0, len(matrix[0])-1
        while up <= down and left <= right:
            for i in range(left, right+1):
                res.append(matrix[up][i])
            up += 1
            if up > down:
                break
            
            for i in range(up, down+1):
                res.append(matrix[i][right])
            right -= 1
            if left > right:
                break
            
            for i in range(right, left-1, -1):
                res.append(matrix[down][i])
            down -= 1
            if up > down:
                break
            
            for i in range(down, up-1, -1):
                res.append(matrix[i][left])
            left += 1
            if left > right:
                break
                
        return res

=================================================================

play with the boundary, break the loop as soon as there is a violation

vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int m = matrix.size();
        if (m == 0) return {};
        int n = matrix[0].size();
        
        vector<int> res;
        int left = 0, right = n - 1, low = m - 1, high = 0;
        while (true) {
            for (int i = left; i <= right; i++) {
                res.push_back(matrix[high][i]);
            }
            if (++high > low)   break;
            
            for (int i = high; i <= low; i++) {
                res.push_back(matrix[i][right]);
            }
            if (--right < left) break;
            
            for (int i = right; i >= left; i--) {
                res.push_back(matrix[low][i]);
            }
            if (--low < high)   break;
            
            for (int i = low; i >= high; i--) {
                res.push_back(matrix[i][left]);
            }
            if (++left > right) break;
        }
        return res;
    }

an implementation like below cannot pass

[[2,3]]

vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int m = matrix.size();
        if (m == 0) return {};
        int n = matrix[0].size();
        
        vector<int> res;
        int left = 0, right = n - 1, low = m - 1, high = 0;
        while (left <= right && high <= low) {
            for (int i = left; i <= right; i++) {
                res.push_back(matrix[high][i]);
            }
            ++high;//if (++high > low)   break;
            
            for (int i = high; i <= low; i++) {
                res.push_back(matrix[i][right]);
            }
            --right;//if (--right < left) break;
            
            for (int i = right; i >= left; i--) {
                res.push_back(matrix[low][i]);
            }
            --low;//if (--low < high)   break;
            
            for (int i = low; i >= high; i--) {
                res.push_back(matrix[i][left]);
            }
            ++right;//if (++left > right) break;
        }
        return res;
    }
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