Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
Example:
Input: 3
Output:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/spiral-matrix-ii
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建立top,bottom,left,right四个边界,创建n*n二维数组并逆时针赋值,当到达边界时,上、左边界自增,右、下边界自减,当top>bottom或left>right时,结束循环。代码如下:
public int[][] generateMatrix(int n) { int[][] res = new int[n][n]; int top = 0; int bottom = n-1; int left = 0; int right = n-1; int target = 1; //看了一下题解,有的大佬把判断循环是否结束的语句设置为target<=n*n while(top <= bottom || left <= right) { //从左往右 for(int i = left; i <= right; i++) { res[top][i] = target++; } top++; //从上往下 for(int i = top; i <= bottom; i++) { res[i][right] = target++; } right--; //从右往左 for(int i = right; i >= left; i--) { res[bottom][i] = target++; } bottom--; //从下往上 for(int i = bottom; i >= top; i--) { res[i][left] = target++; } left++; } return res; }