【细节】剑指offer——面试题20：顺时针打印矩阵

原创已于 2025-04-29 00:46:43 修改 · 434 阅读

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#剑指offer #顺时针打印矩阵

于 2018-03-17 11:42:59 首次发布

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本文提供了一种使用Java实现的顺时针打印矩阵的方法。通过定义边界和循环遍历的方式，实现了对矩阵的有效遍历与打印。代码简洁高效，易于理解。

力扣，https://leetcode.cn/problems/shun-shi-zhen-da-yin-ju-zhen-lcof/description/

Python

class Solution:
    def spiralArray(self, array: List[List[int]]) -> List[int]:
        res = list()
        if len(array) == 0 or len(array[0]) == 0:
            return res
        m, n = len(array), len(array[0])
        m1, n1 = 0, 0
        m2, n2 = m-1, n-1

        while m1 <= m2 and n1 <= n2:
            for i in range(n1, n2+1): # 向右
                res.append(array[m1][i])
            for i in range(m1+1, m2): # 向下
                res.append(array[i][n2])
            for i in range(n2, n1-1, -1): # 向左
                if m1 == m2:
                    break
                res.append(array[m2][i])
            for i in range(m2-1, m1, -1): # 向上
                if n1 == n2:
                    break
                res.append(array[i][n1])
            n1 += 1
            m1 += 1
            n2 -= 1
            m2 -= 1
        
        return res

Java

更新java版，代码更简练，脑瓜子要清晰！！！
注意：横向打印时，索引到两头；纵向打印时，索引只取中间！！！

 class Solution {
    public int[] spiralArray(int[][] array) {
        if (array.length == 0 || array[0].length == 0) {
            return new int[0];
        }
        int xLeftUp = 0, yLeftUp = 0, xRightDown = array.length - 1, yRightDown = array[0].length - 1;
        int[] res = new int[array.length * array[0].length];
        int i = 0;
        while (xLeftUp <= xRightDown && yLeftUp <= yRightDown) {
            for (int k = yLeftUp; k <= yRightDown; ++k) { // 向右
                res[i++] = array[xLeftUp][k];
            }
            for (int k = xLeftUp + 1; k < xRightDown; ++k) { // 向下
                res[i++] = array[k][yRightDown];
            }
            for (int k = yRightDown; k >= yLeftUp && xLeftUp < xRightDown; --k) { // 向左
                res[i++] = array[xRightDown][k];
            }
            for (int k = xRightDown - 1; k > xLeftUp && yLeftUp < yRightDown; --k) { // 向上
                res[i++] = array[k][yLeftUp];
            }
            ++xLeftUp;
            ++yLeftUp;
            --xRightDown;
            --yRightDown;
        }

        return res;
    }
}

Solution1：
可参考leetCode 54题的解法
书上的思路特别好，学习之~

class Solution {
public:
    vector<int> printMatrix(vector<vector<int> > matrix) {
        vector<int> res;
        int rows = matrix.size(), cols = matrix[0].size();
        if(rows == 0 || cols == 0)
            return res;
        int i = 0, j = 0;//循环变量
        while(i <= rows-1-i && j <= cols-1-j){
            PrintCycle(matrix, res, i, j, rows-1, cols-1);
            i++;
            j++;
        }
        return res;
    }
    void PrintCycle(vector<vector<int> > &matrix, vector<int> &result, int m, int n, int row, int col){
        int i = 0, j = 0;
        for(j=n;j<=col-n;j++)
            result.push_back(matrix[m][j]);
        for(i=m+1;i<=row-m;i++)
            result.push_back(matrix[i][col-n]);
        for(j=col-n-1;j>=n && m+1<=row-m;j--)//只有row-m行不与m行重合时才往回push_back
            result.push_back(matrix[row-m][j]);
        for(i=row-m-1;i>m && col-n-1>=n;i--)//只有n行不与col-n行重合时才往上push_back
            result.push_back(matrix[i][n]);
    }
};