[leetcode] Rotate Image

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Follow up:
Could you do this in-place?

https://oj.leetcode.com/submissions/detail/4771943/

思路1：先按主对角线翻转一次，然后左右再翻转一次。每个元素需要两次移动，但是好理解，不易出错。

思路2：直接计算坐标替换，每个元素一次移动即可。

扩展：类似的剑指offer上的 旋转打印矩形内元素的题目，也是坐标操作很麻烦，需利用left，right，up，down四个bar的方法仔细处理。

思路1：

public class Solution {
	public void rotate(int[][] matrix) {
		if (matrix == null)
			return;

		int n = matrix.length;
		int i, j;
		for (i = 0; i < n; i++)
			for (j = 0; j < i; j++) {
				int tmp = matrix[i][j];
				matrix[i][j] = matrix[j][i];
				matrix[j][i] = tmp;
			}

		for (i = 0; i < n; i++)
			for (j = 0; j < n / 2; j++) {
				int tmp = matrix[i][n - 1 - j];
				matrix[i][n - 1 - j] = matrix[i][j];
				matrix[i][j] = tmp;
			}

	}

	public static void main(String[] args) {
		// int[][] matrix = new int[][] { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 }
		// };
		int[][] matrix = new int[][] { { 1, 2, 3, 4 }, { 5, 6, 7, 8 },
				{ 9, 10, 11, 12 }, { 13, 14, 15, 16 } };
		new Solution().rotate(matrix);
	}
}

思路2：

public void rotate2(int[][] matrix) {
        if (matrix == null)
            return;
        printMatrix(matrix);

        for (int i = 0, j = matrix.length - 1; i < j; i++, j--) {
            for (int k = i, d = j; k < j; k++, d--) {
                int t = matrix[i][k];
                matrix[i][k] = matrix[d][i];
                matrix[d][i] = matrix[j][d];
                matrix[j][d] = matrix[k][j];
                matrix[k][j] = t;
            }
        }

[leetcode] Rotate Image,布布扣,bubuko.com

[leetcode] Rotate Image

原文：http://www.cnblogs.com/jdflyfly/p/3810758.html