Force deduction solution summary 498 diagonal traversal

Directory links ：

Force buckle programming problem - The solution sums up _ Share + Record -CSDN Blog

GitHub Synchronous question brushing items ：

https://github.com/September26/java-algorithms

Original link ： Power button

describe ：

Give you a size of m x n Matrix mat , Please traverse diagonally , Use an array to return all the elements in the matrix .

Example 1：

Input ：mat = [[1,2,3],[4,5,6],[7,8,9]]
Output ：[1,2,4,7,5,3,6,8,9]
Example 2：

Input ：mat = [[1,2],[3,4]]
Output ：[1,2,3,4]
 

Tips ：

m == mat.length
n == mat[i].length
1 <= m, n <= 104
1 <= m * n <= 104
-105 <= mat[i][j] <= 105

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/diagonal-traverse
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Their thinking ：

*  Their thinking ：
*  Set both directions , They are upper left and lower right .forward==0 Hour represents upper left ,forward==1 Hour stands for bottom right .
*  Take the upper left search as an example , If (x+1,y-1) Not out of range , Then add this value to the array , otherwise , This indicates that it is time to reverse the search .
*  hold forward Set to 1.

Code ：

public class Solution498 {

    public int[] findDiagonalOrder(int[][] mat) {
        int index = 0;
        int forward = 0;//0 The upper right ,1 The lower left 
        int[] result = new int[mat[0].length * mat.length];
        int x = 0, y = 0;
        while (index < result.length) {
            System.out.println("y:" + y + ",x:" + x);
            result[index++] = mat[y][x];
            if (forward == 0) {
                if ((x + 1) < mat[0].length && (y - 1) >= 0) {
                    x++;
                    y--;
                    continue;
                }
                if ((x + 1) < mat[0].length) {
                    x++;
                } else {
                    y++;
                }
                forward = 1;
                continue;
            }
            if ((x - 1) >= 0 && (y + 1) < mat.length) {
                x--;
                y++;
                continue;
            }
            if (y + 1 < mat.length) {
                y++;
            } else {
                x++;
            }
            forward = 0;
        }
        return result;
    }
}