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.
思路:我们设从二维矩阵右上角那个元素temp = matrix[0][n-1]开始搜索.然后将target与其比较,若和target相等,则返回true;若比target大,则向下搜索;若比target小,则想左搜索。
java代码实现:
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int row = matrix.length;
int col = matrix[0].length;
int i = 0;
int j = col -1;
int temp = matrix[i][j];
while(true){
if(temp == target)
return true;
else if(temp < target && i < row - 1){
temp = matrix[++i][col-1];
}
else if(temp > target && j > 0){
temp = matrix[i][--j];
}
else
return false;
}
}
}