给定一个含有 M x N 个元素的矩阵（M 行，N 列），请以对角线遍历的顺序返回这个矩阵中的所有元素，对角线遍历如下图所示。

示例:

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

输出: [1,2,4,7,5,3,6,8,9]

说明:

1. 给定矩阵中的元素总数不会超过 100000 。

C++1

class Solution {
public:
    vector<int> findDiagonalOrder(vector<vector<int>>& matrix) {
        if (matrix.empty() || matrix[0].empty()) return {};
        int x = matrix.size();
        int y = matrix[0].size();
        vector<int> res(x*y);
        vector<vector<int>> dirs{{-1,1}, {1,-1}};
        
        int r = 0, c = 0, k = 0;
        for(int i=0;i<x*y;i++){
            res[i] = matrix[r][c];
            r+=dirs[k][0];
            c+=dirs[k][1];
            if(r>=x){
                r = x-1;
                c+=2;
                k = 1-k;
            }
            if(c>=y){
                c = y - 1;
                r+=2;
                k = 1-k;
            }
            if(r<0){
                r=0;
                k = 1-k;
            }
            if(c<0){
                c = 0;
                k = 1-k;
            }
        }
        return res;
    }
    
};

C++2

class Solution {
public:
    vector<int> findDiagonalOrder(vector<vector<int>>& matrix) {
        if (matrix.empty() || matrix[0].empty()) return {};
        int m = matrix.size(), n = matrix[0].size(), r = 0, c = 0;
        vector<int> res(m * n);
        for (int i = 0; i < m * n; ++i) {
            res[i] = matrix[r][c];
            if ((r + c) % 2 == 0) {
                if (c == n - 1) {++r;}
                else if (r == 0) {++c;}
                else {--r; ++c;}
            } else {
                if (r == m - 1) {++c;}
                else if (c == 0) {++r;}
                else {++r; --c;}
            }
        }
        return res;
    }
};