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面试题29. 顺时针打印矩阵

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字。

示例 1：

输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,3,6,9,8,7,4,5]
示例 2：

输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
输出：[1,2,3,4,8,12,11,10,9,5,6,7]
 

限制：

0 <= matrix.length <= 100
0 <= matrix[i].length <= 100

/**
 * @param {number[][]} matrix
 * @return {number[]}
 */
/*
// 这种方法有问题，某些特殊测试用例总是会缺失
var spiralOrder = function(matrix) {
    if(!matrix.length) return []
    let h = matrix.length, w = matrix[0].length;
    let res = [];
    for(let i=0; i<h/2; i++){
        for(let j=i; j<w-i; j++){
            res.push(matrix[i][j])
        }
        for(let m=i+1; m<=h-1-i; m++){
            res.push(matrix[m][w-i-1])
        }
        for(let n=w-i-2; n>=0 && i+1<h-1; n--){
            console.log('3: ',matrix[h-i-1][n])
            res.push(matrix[h-i-1][n])
        }
        for(let k=h-i-2; k>=i+1; k--){
            console.log('4: ',matrix[k][i])
            res.push(matrix[k][i])
        }
    }
    return res 
};
*/
var spiralOrder = function(matrix) {
    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return [];
    }
    let rows = matrix.length, columns = matrix[0].length;
    let visited = new Array(rows)
    for(let i=0; i<rows; i++){
        visited[i] = new Array(columns)
    }
    let total = rows * columns;
    let order = new Array(total);
    let row = 0, column = 0;
    let directions = [[0, 1], [1, 0], [0, -1], [-1, 0]];
    let directionIndex = 0;
     for (let i = 0; i < total; i++) {
            order[i] = matrix[row][column];
            visited[row][column] = true;
            let nextRow = row + directions[directionIndex][0], nextColumn = column + directions[directionIndex][1];
            if (nextRow < 0 || nextRow >= rows || nextColumn < 0 || nextColumn >= columns || visited[nextRow][nextColumn]) {
                directionIndex = (directionIndex + 1) % 4;
            }
            row += directions[directionIndex][0];
            column += directions[directionIndex][1];
        }
        return order;
}

思考：

let directions = [[0, 1], [1, 0], [0, -1], [-1, 0]]; 4个方向的题解，不容易出错，好理解，又通用

第一种解法的思路是下面的，控制4个变量往里面缩小距形。

class Solution {
    public int[] spiralOrder(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return new int[0];
        }
        int rows = matrix.length, columns = matrix[0].length;
        int[] order = new int[rows * columns];
        int index = 0;
        int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
        while (left <= right && top <= bottom) {
            for (int column = left; column <= right; column++) {
                order[index++] = matrix[top][column];
            }
            for (int row = top + 1; row <= bottom; row++) {
                order[index++] = matrix[row][right];
            }
            if (left < right && top < bottom) {
                for (int column = right - 1; column > left; column--) {
                    order[index++] = matrix[bottom][column];
                }
                for (int row = bottom; row > top; row--) {
                    order[index++] = matrix[row][left];
                }
            }
            left++;
            right--;
            top++;
            bottom--;
        }
        return order;
    }
}

　　

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof
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