leetcode 240 Search a 2D Matrix II

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本文介绍了一种高效的矩阵搜索算法，该算法能在排序的二维矩阵中查找特定的目标值。矩阵的每一行从左到右递增排序，每一列从上到下递增排序。通过递归划分矩阵，逐步缩小搜索范围来提高查找效率。

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.

For 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.

public class Solution {
    public boolean search(int[][] matrix,int a, int b, int m, int n, int target) {

        if(a > m || b > n) return false;

        if(matrix[m][n] < target || matrix[a][b] > target) return false;

        else if(matrix[m][n] == target || matrix[a][b] == target) return true;

        else {
            int left_top_row = a, left_top_column = b, right_buttom_row = m, right_buttom_column = n;

            int low = left_top_column, high = right_buttom_column;
            int row = left_top_row;

            while (low <= high) {
                int middle = (low + high)/2;

                if (matrix[row][middle] == target) return true;
                else if (matrix[row][middle] > target) {
                    high = middle-1;
                } else {
                    low = middle+1;
                }
            }

            if(low < right_buttom_column) {
                n = low;
            } else {
                a += 1;
            }

            low = left_top_row;
            high = right_buttom_row;
            int column = left_top_column;

            while (low <= high) {
                int middle = (low+high)/2;

                if (matrix[middle][column] == target) return true;
                else if(matrix[middle][column] > target) high = middle-1;
                else low = middle + 1;
            }

            if(low < right_buttom_row) {
                m = low;
            } else {
                b += 1;
            }

            return search(matrix, a, b, m, n, target);
        }
    }
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length, n=matrix[0].length;


        return search(matrix,0,0,m-1,n-1,target);
    }
}