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.

M1: search from the bottom left corner

观察到，左下角的数字，向上递减，向右递增，可以从左下角开始查找，循环成立条件是数组下标不越界。如果 当前数 < target，向右查找；如果 当前数 < target，向上查找。

time: O(m + n), space: O(1)

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0) {
            return false;
        }
        int row = matrix.length - 1, col = 0;
        while(row >= 0 && col <= matrix[0].length - 1) {
            if(matrix[row][col] == target) {
                return true;
            } else if(matrix[row][col] > target) {
                row--;
            } else {
                col++;
            }
        }
        return false;
    }
}

M2: 利用 recursion + binary search

The basic idea is to use recursion and a helper function (search). In each recursive call, check the middle element if it's equal to target. If it's equal, return true as result, otherwise keep searching on the remaining sub-matrix where the target could still possibly in by recursively calling the search function. Here if middle element is not equal to target, we can exclude an area that the target cannot be in, and the remaining submatrix is like an L shaped area, which can be split into two matrices and search respectively (if any matrix exists the target, return true).

time =  O(logm + logn), space = O(logmn)

class Solution {
    int[][] matrix;
    int target;
    
    public boolean searchMatrix(int[][] matrix, int target) {
        this.matrix = matrix;
        this.target = target;
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }
        return search(0, 0, matrix.length - 1, matrix[0].length - 1);
    }
    
    public boolean search(int x1, int y1, int x2, int y2) {
        if(x1 > x2 || y1 > y2 || x1 > matrix.length - 1 || x2 < 0 || y1 > matrix[0].length - 1 || y2 < 0) {
            return false;
        }
        int midx = x1 + (x2 - x1) / 2;
        int midy = y1 + (y2 - y1) / 2;
        if(matrix[midx][midy] == target) {
            return true;
        } else if(matrix[midx][midy] < target) {
            return search(x1, midy + 1, x2, y2) || search(midx + 1, y1, x2, midy);
        } else if(matrix[midx][midy] > target) {
            return search(x1, y1, x2, midy - 1) || search(x1, midy, midx - 1, y2);
        } else {
            return false;
        }
    }
}