题目链接：search-a-2d-matrix
/**
*
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.
*
*/
public class SearchA2DMatrix {
// 134 / 134 test cases passed.
// Status: Accepted
// Runtime: 236 ms
// Submitted: 0 minutes ago
//时间复杂度 O(log(max(m,n)) 空间复杂度 O(1)
static boolean searchMatrix(int[][] matrix, int target) {
int m = matrix.length; //行数
int n = matrix[0].length; //列数
int up = 0;
int down = m - 1;
int left = 0;
int right = n;
//二分法查找所在行
while(up < down) {
int mid = (down + up) / 2;
if(matrix[mid][n - 1] > target) down = mid;
else if(matrix[mid][n - 1] < target) up = mid + 1;
else return true;
}
//二分法查找所在列
while(left < right) {
int mid = (left + right) / 2;
if(matrix[up][mid] > target) right = mid;
else if(matrix[up][mid] < target) left = mid + 1;
else return true;
}
return false;
}
public static void main(String[] args) {
System.out.println(searchMatrix(new int[][]{{1, 3, 5, 7},
{10, 11, 16, 20},
{23, 30, 34, 50}}, 3));
System.out.println(searchMatrix(new int[][]{{1, 1}}, 2));
System.out.println(searchMatrix(new int[][]{{1,}, { 3}}, 2));
System.out.println(searchMatrix(new int[][]{{1}}, 1));
}
}
[LeetCode 74] Search a 2D Matrix
原文：http://blog.csdn.net/ever223/article/details/44514849