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
.
思路:
这道题非常有技巧,我们选择右上角的那个点,然后进行比较,如果大,则向下走,如果小则向右。(因为从那个点开始,下面永远大于,左边永远小于)
class Solution { public boolean searchMatrix(int[][] matrix, int target) { if(matrix.length ==0) return false; if(matrix[0].length ==0) return false; int row =0; int col = matrix[0].length -1; while(row<matrix.length && col >=0){ if(matrix[row][col] == target){ return true; } if(matrix[row][col] > target){ col --; } else if(matrix[row][col] < target){ row++; } } return false; } }