題目：

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：

二分查找

package array;public class SearchA2DMatrix {public boolean searchMatrix(int[][] matrix, int target) {int m;int n;if (matrix == null || (m = matrix.length) == 0 || (n = matrix[0].length) == 0) return false;int start = 0;int end = m * n - 1;while (start <= end) {int mid = (end - start) / 2 + start;int i = mid / n;int j = mid - i * n;if (matrix[i][j] == target)return true;if (matrix[i][j] < target)start = mid + 1;elseend = mid - 1;}return false;}public static void main(String[] args) {// TODO Auto-generated method stubint[][] matrix = {{ 1, 3, 5, 7 },{ 10, 11, 16, 20 },{ 23, 30, 34, 50}};SearchA2DMatrix s = new SearchA2DMatrix();System.out.println(s.searchMatrix(matrix, 8));}}

?

轉載于:https://www.cnblogs.com/null00/p/5093203.html

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