力扣---2020.4.15

542. 01 矩陣

//動態規劃
class Solution {
  public int[][] updateMatrix(int[][] matrix) {
    int m = matrix.length, n = matrix[0].length;
    int[][] dp = new int[m][n];
    for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {
        dp[i][j] = matrix[i][j] == 0 ? 0 : 10000;
      }
    }

    // 從左上角開始
    for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {
        if (i - 1 >= 0) {
          dp[i][j] = Math.min(dp[i][j], dp[i - 1][j] + 1);
        }
        if (j - 1 >= 0) {
          dp[i][j] = Math.min(dp[i][j], dp[i][j - 1] + 1);
        }
      }
    }
    // 從右下角開始
    for (int i = m - 1; i >= 0; i--) {
      for (int j = n - 1; j >= 0; j--) {
        if (i + 1 < m) {
          dp[i][j] = Math.min(dp[i][j], dp[i + 1][j] + 1);
        }
        if (j + 1 < n) {
          dp[i][j] = Math.min(dp[i][j], dp[i][j + 1] + 1);
        }
      }
    }
    return dp;
  }
}

//BFS
class Solution {
    public int[][] updateMatrix(int[][] matrix) {
        // 首先將所有的 0 都入隊，並且將 1 的位置設置成 -1，表示該位置是 未被訪問過的 1
        Queue<int[]> queue = new LinkedList<>();
        int m = matrix.length, n = matrix[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == 0) {
                    queue.offer(new int[] {i, j});
                } else {
                    matrix[i][j] = -1;
                } 
            }
        }
        
        int[] dx = new int[] {-1, 1, 0, 0};
        int[] dy = new int[] {0, 0, -1, 1};
        while (!queue.isEmpty()) {
            int[] point = queue.poll();
            int x = point[0], y = point[1];
            for (int i = 0; i < 4; i++) {
                int newX = x + dx[i];
                int newY = y + dy[i];
                // 如果四鄰域的點是 -1，表示這個點是未被訪問過的 1
                // 所以這個點到 0 的距離就可以更新成 matrix[x][y] + 1。
                if (newX >= 0 && newX < m && newY >= 0 && newY < n 
                        && matrix[newX][newY] == -1) {
                    matrix[newX][newY] = matrix[x][y] + 1;
                    queue.offer(new int[] {newX, newY});
                }
            }
        }

        return matrix;
    }
}

面試題68 - II. 二叉樹的最近公共祖先

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root==null){
            return null;
        }
        if(root==p || root==q){
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left,p,q);
        TreeNode right = lowestCommonAncestor(root.right,p,q);
        if(left!=null&&right!=null){
            return root;
        }
        if(left!=null){
            return left;
        }
        if(right!=null){
            return right;
        }
        return null;
    }
}