给你一个 m 行 n 列的矩阵 matrix ,请按照 顺时针螺旋顺序 ,返回矩阵中的所有元素。
示例一:
输入:matrix = [[1,2,3],[4,5,6],[7,8,9]]输出:[1,2,3,6,9,8,7,4,5]
实例二:
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]
提示:
m == matrix.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100
题解
此题和59一样,模拟就行了,没啥难的。注意边界的判断!
JavaScript
/*** @param {number[][]} matrix* @return {number[]}*/var spiralOrder = function(matrix) {ans = []let top = 0let left = 0let right = matrix[0].length-1let bottom = matrix.length-1let num = 1let target = matrix[0].length*matrix.lengthwhile(num<=target){for(let i=left;i<right+1;i++){if(num>target) continueans.push(matrix[top][i])num++}top++for(let i=top;i<bottom+1;i++){if(num>target) continueans.push(matrix[i][right])num++}right--for(let i=right;i>left-1;i--){if(num>target) continueans.push(matrix[bottom][i])num++}bottom--for(let i=bottom;i>top-1;i--){if(num>target) continueans.push(matrix[i][left])num++}left++}return ans};let m = [[1,2,3],[4,5,6],[7,8,9]]let matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]console.log(spiralOrder(matrix))
