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.Example 1:
Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
Output: true
Example 2:
Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
Output: false
难度：medium
题目：写算法在 m * n 的矩阵中查找给定的值。矩阵特征如下：矩阵中的每行升序排列。每行中的的第一个数大于其前一行的最后一个数。
思路：从最后一列中找出第一个大于 target 的值，并记录下当前行。然后对该行进行二分查找。
Runtime: 5 ms, faster than 73.46% of Java online submissions for Search a 2D Matrix.Memory Usage: 38.8 MB, less than 0.96% of Java online submissions for Search a 2D Matrix.
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix.length == 0 || matrix[0].length == 0) {
return false;
}
int m = matrix.length, n = matrix[0].length, row = 0;
for (int i = 0; i < m; i++) {
row = i;
if (matrix[i][n – 1] >= target) {
break;
}
}
return Arrays.binarySearch(matrix[row], target) >= 0;
}
}