498. Diagonal Traverse - Medium - Leetcode - Coding | Hsing Yu = = We are the dreamers. We make it happen 'cause we can see it.

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# 題目敘述

Given an m x n matrix mat , return an array of all the elements of the array in a diagonal order.

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

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

	import java.util.ArrayList;

	class Solution {
	public int[] findDiagonalOrder(int[][] mat) {
	if (mat == null \|\| mat.length == 0) {
	return new int[0];
	}

	int row = mat.length;
	int col = mat[0].length;
	ArrayList<Integer> ans = new ArrayList<>(row * col);

	for (int i = 0; i <= (row + col - 2); i++) {
	int r = (i < col) ? 0 : i - col + 1;
	int c = (i < col) ? i : col - 1;
	ArrayList<Integer> temp = new ArrayList<>();

	while (r < row && c >= 0) {
	temp.add(mat[r][c]);
	r++;
	c--;
	}

	if ((r + c) % 2 == 0) {
	// move up
	for (int j = temp.size() - 1; j >= 0; j--) {
	ans.add(temp.get(j));
	}
	} else {
	// move down
	ans.addAll(temp);
	}
	}

	return ans.stream().mapToInt(i -> i).toArray();
	}
	}

	#include <vector>
	#include <algorithm>

	class Solution {
	public:
	std::vector<int> findDiagonalOrder(std::vector<std::vector<int>>& mat) {
	int row = mat.size();
	int col = mat[0].size();
	std::vector<int> result;

	for(int i = 0; i < row + col - 1; i++) {
	int r = (i < col) ? 0 : i - col + 1;
	int c = (i < col) ? i : col - 1;
	std::vector<int> temp;

	while(r < row && c >= 0) {
	temp.push_back(mat[r][c]);
	r++;
	c--;
	}

	if((r + c) % 2 == 0) {
	std::reverse(temp.begin(), temp.end());
	}
	result.insert(result.end(), temp.begin(), temp.end());
	}

	return result;
	}
	};

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