数组练习-螺旋矩阵

分析

横向纵向每次取得的长度减1，横向或纵向每次方向会调换，一次下标加1，一次减1

虽然看了别人的是回字递归。。。啊这

代码

/**
 * 
 * @param matrix int整型二维数组 
 * @return int整型一维数组
 */
function spiralOrder( matrix ) {
    // write code here
    if (matrix.length === 0) return matrix
    if (matrix.length === 1) return matrix[0]
    
    let m = matrix.length - 1
    let n = matrix[0].length

    let ans = []

    let currenti = 0, currentj = -1
    let arrow = true

    while(m > 0 || n > 0){
        for (let i = 0; i < n; i ++){
            if (arrow) currentj += 1
            else currentj -= 1

            ans.push(matrix[currenti][currentj])
        }
        -- n

        for (let j = 0; j < m; j++){
            if (arrow) currenti += 1
            else currenti -= 1

            ans.push(matrix[currenti][currentj])
        }
        -- m
        arrow = !arrow
    }

    return ans.slice(0,matrix.flat().length)   
}