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
.
分析:
写出一个高效的算法来搜索 m × n矩阵中的值。这个矩阵具有以下特性:
- 每行中的整数从左到右是排序的。
- 每行的第一个数大于上一行的最后一个整数。
这道题可以把二维数组里的数字当作是一个有序的列表,用二分法查找。
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0) {
return false;
}
if (matrix[0] == null || matrix[0].length == 0) {
return false;
} int row = matrix.length, column = matrix[0].length;
int start = 0, end = row * column - 1; while (start + 1 < end) {
int mid = start + (end - start) / 2;
int number = matrix[mid / column][mid % column];
if (number == target) {
return true;
} else if (number < target) {
start = mid;
} else {
end = mid;
}
} if (matrix[start / column][start % column] == target) {
return true;
} else if (matrix[end / column][end % column] == target) {
return true;
} return false; }
}