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  • [LeetCode] 59. Spiral Matrix II 螺旋矩阵 II

    Given an integer n, generate a square matrix filled with elements from 1 to n^2 in spiral order.

    For example,
    Given n = 3,

    You should return the following matrix:

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

    54. Spiral Matrix 类似,这次给定一个整数n,以螺旋顺序填充元素到n^2的矩阵。

    Java:

    class Solution {
        public int[][] generateMatrix(int n) {
            int[][] res = new int[n][n];
            int k = 1;
            int top = 0, bottom = n - 1, left = 0, right = n - 1;
            while (left < right && top < bottom) {
                for (int j = left; j < right; j++) {
                    res[top][j] = k++;
                }
                for (int i = top; i < bottom; i++) {
                    res[i][right] = k++;
                }
                for (int j = right; j > left; j--) {
                    res[bottom][j] = k++;
                }
                for (int i = bottom; i > top; i--) {
                    res[i][left] = k++;
                }
                left++;
                right--;
                top++;
                bottom--;
            }
            if (n % 2 != 0)
                res[n / 2][n / 2] = k;
            return res;
        }
    }
    

    Python:

    class Solution:
        # @param matrix, a list of lists of integers
        # @return a list of integers
        def spiralOrder(self, matrix):
            result = []
            if matrix == []:
                return result
            
            left, right, top, bottom = 0, len(matrix[0]) - 1, 0, len(matrix) - 1
            
            while left <= right and top <= bottom:
                for j in xrange(left, right + 1):
                    result.append(matrix[top][j])
                for i in xrange(top + 1, bottom):
                    result.append(matrix[i][right])
                for j in reversed(xrange(left, right + 1)):
                    if top < bottom:
                        result.append(matrix[bottom][j])
                for i in reversed(xrange(top + 1, bottom)):
                    if left < right:
                        result.append(matrix[i][left])
                left, right, top, bottom = left + 1, right - 1, top + 1, bottom - 1
                
            return result 

    C++:

    class Solution {
    public:
        vector<vector<int> > generateMatrix(int n) {
            vector<vector<int> > res(n, vector<int>(n, 1));
            int val = 1, p = n;
            for (int i = 0; i < n / 2; ++i, p -= 2) {
                for (int col = i; col < i + p; ++col)
                    res[i][col] = val++;
                for (int row = i + 1; row < i + p; ++row)
                    res[row][i + p - 1] = val++;
                for (int col = i + p - 2; col >= i; --col)
                    res[i + p - 1][col] = val++;
                for (int row = i + p - 2; row > i; --row)    
                    res[row][i] = val++;
            }
            if (n % 2 != 0) res[n / 2][n / 2] = val;
            return res;
        }
    };
    

      

    类似题目:

    [LeetCode] 54. Spiral Matrix 螺旋矩阵

    All LeetCode Questions List 题目汇总

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  • 原文地址:https://www.cnblogs.com/lightwindy/p/8564389.html
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