74. Search a 2D Matrix

摘要： 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 ar...

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

難度：medium

題目：寫演算法在m * n的矩陣中查詢給定的值。矩陣特徵如下：

矩陣中的每行升序排列。

每行中的的第一個數大於其前一行的最後一個數。

思路：從最後一列中找出第一個大於target的值，並記錄下當前行。然後對該行進行二分查詢。

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;
}
}