Spiral Matrix

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

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

You should return [1,2,3,6,9,8,7,4,5].

思路：这是个m,n矩阵，每次绕着外围扫一圈，扫到剩下的最后一个 ［ x ］， ［x, y, z］, [ x ], [ y ]

那么定义宽和高，可得到四个角。扫完第一行，heightStart+1，扫完最右边一列，widthEnd-1。

由于heightStart加了一，check是否已经是最底下一行。 

var spiralOrder = function(matrix) {    var result = [];    var n = matrix.length;    var m = (matrix.length===0? 0:matrix[0].length);    var widthStart = 0,        widthEnd = m-1,        heightStart = 0,        heightEnd = n-1;    while (widthStart<=widthEnd&&heightStart<=heightEnd) {        for (var i=widthStart;i<=widthEnd;i++) {            result.push(matrix[heightStart][i]);        }        heightStart++        for (var j=heightStart;j<=heightEnd;j++) {            result.push(matrix[j][widthEnd]);        }        widthEnd--        if (heightStart<=heightEnd) {            for (var k=widthEnd;k>=widthStart;k--) {               result.push(matrix[heightEnd][k]);            }                  heightEnd--        }        if (widthStart<=widthEnd) {            for (var x=heightEnd;x>=heightStart;x--) {                result.push(matrix[x][widthStart])            }            widthStart++        }    }        return result;};