https://leetcode.com/problems/spiral-matrix/

开始还想用递归来遍历输出，但是需要对数组进行裁剪，处理麻烦而且占用空间。
遇到这种map问题可以用变量来标志范围。

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/**
 * @param {number[][]} matrix
 * @return {number[]}
 */
var spiralOrder = function(matrix) {
    let ret = []
    if (matrix.length ===0) return ret
    let rowBegin = 0
    let rowEnd = matrix.length - 1
    let colBegin = 0
    let colEnd = matrix[0].length - 1
    while(rowBegin <= rowEnd && colBegin <= colEnd) {
        for (let i = colBegin; i <= colEnd; i++) {
            ret.push(matrix[rowBegin][i])
        }
        rowBegin++
        for (let i = rowBegin; i <= rowEnd; i++) {
            ret.push(matrix[i][colEnd])
        }
        colEnd--
        if (rowBegin <= rowEnd) {
            for (let i = colEnd; i >= colBegin; i--) {
                ret.push(matrix[rowEnd][i])
            }
        }
        rowEnd--
        if (colBegin <= colEnd) {
            for (let i = rowEnd; i >= rowBegin; i--) {
                ret.push(matrix[i][colBegin])
            }
        }
        colBegin++
    }
    return ret
};