0498. Diagonal Traverse (M)

Diagonal Traverse (M)

题目

Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order as shown in the below image.

Example:

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

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

Explanation:

Note:

The total number of elements of the given matrix will not exceed 10,000.

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题意

按照对角线方向遍历矩阵。

思路

可以按照图示路线进行模拟，或者直接每次都按照右上到左下的顺序遍历，并把奇数次的遍历翻转过来：

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代码实现

Java

模拟

class Solution {
    public int[] findDiagonalOrder(int[][] matrix) {
        if (matrix.length == 0) {
            return new int[0];
        }

        int m = matrix.length, n = matrix[0].length;
        int[] ans = new int[m * n];
        int x = 0, y = 0;
        int xStep = -1, yStep = 1;

        for (int i = 0; i < m * n; i++) {
            ans[i] = matrix[x][y];
            int nextX = x + xStep, nextY = y + yStep;

            if (nextX == -1 && nextY == n) {
              	// 当前在右上角，且要继续向右上走
                x = x + 1;
            } else if (nextX == -1 || nextY == n) {
              	// 当前在上边界或右边界，且要继续向右上走
                x = nextX == -1 ? x : x + 1;
                y = nextY == n ? y : y + 1;
            } else if (nextX == m && nextY == -1) {
              	// 当前在左下角，且要继续向左下走
                y = y + 1;
            } else if (nextX == m || nextY == -1) {
              	// 当前在左边界或下边界，且要继续向左下走
                x = nextX == m ? x : x + 1;
                y = nextY == -1 ? y : y + 1;
            } else {
                x = nextX;
                y = nextY;
            }

            if (nextX == -1 || nextY == -1 || nextX == m || nextY == n) {
                xStep = -xStep;
                yStep = -yStep;
            }
        }

        return ans;
    }
}

翻转

class Solution {
    public int[] findDiagonalOrder(int[][] matrix) {
        if (matrix.length == 0) {
            return new int[0];
        }

        int m = matrix.length, n = matrix[0].length;
        int[] ans = new int[m * n];
        int index = 0;

        for (int i = 0; i < m + n - 1; i++) {
            int x = i < n ? 0 : i - n + 1, y = i < n ? i : n - 1;
            int start = index;
            while (x < m && y >= 0) {
                ans[index++] = matrix[x++][y--];
            }
            if (i % 2 == 0) {
                reverse(ans, start, index - 1);
            }
        }

        return ans;
    }

    private void reverse(int[] A, int start, int end) {
        while (start < end) {
            int tmp = A[start];
            A[start++] = A[end];
            A[end--] = tmp;
        }
    }
}