Question

498. Diagonal Traverse

中文文档

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

Solution 1

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

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;
  }
};

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()
  }
}