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:先利用每一行的最后一个数与target判断,确定target可能在的行数;再在确定的那一行进行折半查找。
public class S074 {
public boolean searchMatrix(int[][] matrix, int target) {
int i = 0;
for (;i<matrix.length;i++) {
if (matrix[i][matrix[0].length-1]>=target) {
break;
}
}
if (i == matrix.length)
return false;
int l = 0, r = matrix[0].length-1;
//折半查找
while (l <= r) {
if (target == matrix[i][(l+r)/2])
return true;
else if (target > matrix[i][(l+r)/2])
l = (l+r)/2+1;
else
r = (l+r)/2-1;
}
return false;
}
}