【LeetCode】74. Search a 2D Matrix

1.题目描述

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 from left to right.
The first integer of each row is greater than the last integer of the previous row.

编写一个高效的算法来判断 m x n 矩阵中，是否存在一个目标值。该矩阵具有如下特性：

每行中的整数从左到右按升序排列。
每行的第一个整数大于前一行的最后一个整数。

Example 1:

Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
Output: false

2.Solutions

Use binary search.

n m matrix convert to an array => matrix[x][y] => a[x m + y] m为一列的长度

an array convert to n * m matrix => a[x] =>matrix[x / m][x % m];

public boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length == 0)
        return false;

    int start = 0, rows = matrix.length, cols = matrix[0].length;
    int end = rows * cols - 1;

    while (start <= end) {
        int mid = start + (end - start >> 1);
        if (matrix[mid / cols][mid % cols] == target) {
            return true;
        } 
        if (matrix[mid / cols][mid % cols] < target) {
            start = mid + 1;
        } else {
            end = mid - 1;
        }
    }
    return false;
}

（完）