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498. Diagonal Traverse

Description

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

 

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]

 

Constraints:

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

Solutions

  • class Solution {
        public int[] findDiagonalOrder(int[][] mat) {
            int m = mat.length, n = mat[0].length;
            int[] ans = new int[m * n];
            int idx = 0;
            List<Integer> t = new ArrayList<>();
            for (int k = 0; k < m + n - 1; ++k) {
                int i = k < n ? 0 : k - n + 1;
                int j = k < n ? k : n - 1;
                while (i < m && j >= 0) {
                    t.add(mat[i][j]);
                    ++i;
                    --j;
                }
                if (k % 2 == 0) {
                    Collections.reverse(t);
                }
                for (int v : t) {
                    ans[idx++] = v;
                }
                t.clear();
            }
            return ans;
        }
    }
    
  • class Solution {
    public:
        vector<int> findDiagonalOrder(vector<vector<int>>& mat) {
            int m = mat.size(), n = mat[0].size();
            vector<int> ans;
            vector<int> t;
            for (int k = 0; k < m + n - 1; ++k) {
                int i = k < n ? 0 : k - n + 1;
                int j = k < n ? k : n - 1;
                while (i < m && j >= 0) t.push_back(mat[i++][j--]);
                if (k % 2 == 0) reverse(t.begin(), t.end());
                for (int& v : t) ans.push_back(v);
                t.clear();
            }
            return ans;
        }
    };
    
  • class Solution:
        def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
            m, n = len(mat), len(mat[0])
            ans = []
            for k in range(m + n - 1):
                t = []
                i = 0 if k < n else k - n + 1
                j = k if k < n else n - 1
                while i < m and j >= 0:
                    t.append(mat[i][j])
                    i += 1
                    j -= 1
                if k % 2 == 0:
                    t = t[::-1]
                ans.extend(t)
            return ans
    
    
  • func findDiagonalOrder(mat [][]int) []int {
    	m, n := len(mat), len(mat[0])
    	var ans []int
    	for k := 0; k < m+n-1; k++ {
    		var t []int
    		i, j := k-n+1, n-1
    		if k < n {
    			i, j = 0, k
    		}
    		for i < m && j >= 0 {
    			t = append(t, mat[i][j])
    			i++
    			j--
    		}
    		if k%2 == 0 {
    			p, q := 0, len(t)-1
    			for p < q {
    				t[p], t[q] = t[q], t[p]
    				p++
    				q--
    			}
    		}
    		for _, v := range t {
    			ans = append(ans, v)
    		}
    	}
    	return ans
    }
    
  • function findDiagonalOrder(mat: number[][]): number[] {
        const res = [];
        const m = mat.length;
        const n = mat[0].length;
        let i = 0;
        let j = 0;
        let mark = true;
        while (res.length !== n * m) {
            if (mark) {
                while (i >= 0 && j < n) {
                    res.push(mat[i][j]);
                    i--;
                    j++;
                }
                if (j === n) {
                    j--;
                    i++;
                }
                i++;
            } else {
                while (i < m && j >= 0) {
                    res.push(mat[i][j]);
                    i++;
                    j--;
                }
                if (i === m) {
                    i--;
                    j++;
                }
                j++;
            }
            mark = !mark;
        }
        return res;
    }
    
    
  • impl Solution {
        pub fn find_diagonal_order(mat: Vec<Vec<i32>>) -> Vec<i32> {
            let (m, n) = (mat.len(), mat[0].len());
            let (mut i, mut j) = (0, 0);
            (0..m * n)
                .map(|_| {
                    let res = mat[i][j];
                    if (i + j) % 2 == 0 {
                        if j == n - 1 {
                            i += 1;
                        } else if i == 0 {
                            j += 1;
                        } else {
                            i -= 1;
                            j += 1;
                        }
                    } else {
                        if i == m - 1 {
                            j += 1;
                        } else if j == 0 {
                            i += 1;
                        } else {
                            i += 1;
                            j -= 1;
                        }
                    }
                    res
                })
                .collect()
        }
    }
    
    

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