【LeetCode】Spiral Matrix II

最新推荐文章于 2019-05-14 09:57:51 发布

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本文介绍了一种生成螺旋矩阵的方法，即按螺旋顺序填充从1到n²的元素到一个n×n的矩阵中。提供了Java和Python两种语言的实现方案，并详细解释了通过设定边界值来完成矩阵填充的过程。

Spiral Matrix II
Total Accepted: 9622 Total Submissions: 31755 My Submissions
Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
For example,
Given n = 3 ,
You should return the following matrix:

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

【解题思路】
和 Spiral Matrix一样，思路参考【LeetCode】Spiral Matrix，但是这次不是打印矩阵的内容了，而是赋值，思路仍然一致。
1、设定边界值，startx，endx，starty，endy。
2、左到右，上到下，右到左，下到上，直到while条件结束。

Java AC 348MS

public class Solution {
    public int[][] generateMatrix(int n) {
		int [][]matrix = new int[n][n];
		int startx = 0;
		int starty = 0;
		int endx = n - 1;
		int endy = n - 1;
		int num = 1;
		while (startx <= endx && starty <= endy) {
			for (int i = starty; i <= endy; i++) {
				matrix[startx][i] = num;
				num++;
			}
			for (int i = startx+1; i <= endx; i++) {
				matrix[i][endy] = num;
				num++;
			}
			if (startx != endx) {
				for (int i = endy-1; i >= starty; i--) {
					matrix[endx][i] = num;
					num++;
				}
			}
			if (starty != endy) {
				for (int i = endx-1; i > startx; i--) {
					matrix[i][starty] = num;
					num++;
				}
			}
			startx++;
			endx--;
			starty++;
			endy--;
		}
		return matrix;
	}
}

Python AC 208ms

class Solution:
    # @return a list of lists of integer
    def generateMatrix(self, n):
        matrix = [[0 for col in range(n)] for row in range(n)]
        startx = 0
        starty = 0
        endx = n - 1
        endy = n - 1
        num = 1
        while startx <= endx and starty <= endy:
            for i in range(starty, endy + 1):
                matrix[startx][i] = num
                num += 1
            for i in range(startx + 1, endx + 1):
                matrix[i][endy] = num
                num += 1
            if startx != endx:
                for i in range(endy - 1, starty - 1, -1):
                    matrix[endx][i] = num
                    num += 1
            if starty != endy:
                for i in range(endx - 1, startx, -1):
                   matrix[i][starty] = num
                   num += 1
            startx += 1
            endx -= 1
            starty += 1
            endy -= 1
        return matrix