240. Search a 2D Matrix II

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

    Integers in each row are sorted in ascending from left to right.
    Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.
  • Divide And Conquer

public boolean searchMatrix(int[][] matrix, int target) {
    if(matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0) return false;
    return divide(matrix, target, 0, matrix[0].length-1, 0, matrix.length-1);
}
public boolean divide(int[][] matrix, int target, int c1, int c2, int r1, int r2) {
    if(c2 < c1 || r2 < r1) return false;
    int c = c1 + (c2-c1) / 2, r = r1 + (r2-r1) / 2;
    if(matrix[r][c] == target) return true;
    else if(matrix[r][c] > target) {
        return divide(matrix, target, c1, c-1, r1, r-1) || divide(matrix, target, c, c2, r1, r-1) || divide(matrix, target, c1, c-1, r, r2);
    } else {
        return divide(matrix, target, c+1, c2, r+1, r2) || divide(matrix, target, c+1, c2, r1, r) || divide(matrix, target, c1, c, r+1, r2);
    }
}
  • TwoPointers
public boolean searchMatrix(int[][] matrix, int target) {
    if(matrix.length == 0 || matrix[0].length == 0) return false;
    int i = 0, j = matrix[0].length-1;
    while(i < matrix.length && j >= 0) {
        if(matrix[i][j] == target) return true;
        else if(matrix[i][j] < target) i++;
        else j--;
    }
    return false;
}