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【leetcode】Search a 2D Matrix

2024-07-08 22:09:50   231人阅读

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.

 

For example,

Consider the following matrix:

[  [1,   3,  5,  7],  [10, 11, 16, 20],  [23, 30, 34, 50]]

Given target = 3, return true

题解：直接把二维数组看成一个长度为m*n的数列，然后用二分法查找target，注意在这个数列中编号为k的数在二维数组中的坐标x = k/n, y=k%n;

代码如下;

 1 public class Solution { 2     public boolean searchMatrix(int[][] matrix, int target) { 3         if(matrix.length == 0 || matrix[0].length == 0) 4             return false; 5         int m = matrix.length; 6         int n = matrix[0].length; 7          8         int start = 0,end = m*n-1; 9         while(start <= end){10             int mid = start + (end-start)/2;11             int x = mid/n;12             int y = mid%n;13             14             if(matrix[x][y] == target)15                 return true;16             17             if(matrix[x][y] < target)18                 start = mid+1;19             else20                 end = mid-1;21         }22         return false;23     }24 }

 

 


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