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.
解法一：暴力解法，遍历二维数组。
代码如下：
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int rows=matrix.length;
int columns=matrix[0].length;
for(int i=0;i<rows;i++)
for(int j=0;j<columns;j++){
if(target==matrix[i][j])
return true;
}
return false;
}
}
运行结果：时间复杂度为O(M*N)
方法二：根据矩阵的特点，从右上角元素开始比较，target小，则列数减1，target大，则行数加1.
代码如下：
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix==null ||matrix.length<1 || matrix[0].length<1){
return false;
}
int col=matrix[0].length-1;
int row=0;
while(col>=0 && row<=matrix.length-1){
if(target==matrix[row][col]){
return true;
}else if(target<matrix[row][col]){
col--;
}else if(target>matrix[row][col]){
row++;
}
}
return false;
}
}
运行结果：时间复杂度O(M+N)
