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LeetCode-54-Spiral Matrix

end row result || ive res () 算法 mce

算法描述:

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

Input:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]

Example 2:

Input:
[

[1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

解題思路:模擬方法解決,註意邊界及細節。

    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> result;
        if(matrix.size()==0 || matrix[0].size() == 0)
            return result;
        int rowBegin = 0;
        int rowEnd = matrix.size()-1;
        int colBegin = 0
; int colEnd = matrix[0].size()-1; while(rowBegin <= rowEnd && colBegin <= colEnd){ for(int i=colBegin; i<= colEnd; i++) result.push_back(matrix[rowBegin][i]); rowBegin++; for(int i=rowBegin; i<= rowEnd; i++) result.push_back(matrix[i][colEnd]); colEnd
--; if(colBegin <= colEnd && rowBegin <= rowEnd){ for(int i = colEnd; i >= colBegin; i--) result.push_back(matrix[rowEnd][i]); rowEnd--; for(int i = rowEnd; i >= rowBegin; i--) result.push_back(matrix[i][colBegin]); colBegin++; } } return result; }

LeetCode-54-Spiral Matrix