Search a 2D Matrix

Description:

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.

问题描述：

给定矩阵，从左到右，从上到下，元素逐渐增大。

解法一：

思路：

基本二分法思路，考点在于把矩阵元素铺开成数组之间的转换关系（按列铺开）

Code:

public class Solution { 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)/ 2; 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; } }