Difficulty:medium
More:【目录】LeetCode Java实现
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 in ascending from left to right.
- Integers in each column are sorted in ascending from top to bottom.
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.
Intuition
refer to 二维数组中的查找
Solution
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix==null || matrix.length<=0 || 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;
else if(matrix[row][col]>target)
col--;
else if(matrix[row][col]<target)
row++;
}
return false;
}
Complexity
Time complexity : O(m+n)
Space complexity : O(1)
What I've learned
1. Ought to have a good command of the thought in this problem.
More:【目录】LeetCode Java实现