题目：
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.
链接： http://leetcode.com/problems/search-a-2d-matrix/
题解：
可以把矩阵当做一个长的行向量进行binary search， 也可以对行列分别进行两次binary search，都差不多。
Time Complexity - O(log(mn))， Space Complexity - O(1).
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix == null || matrix.length == 0)
return false;
int rowNum = matrix.length, colNum = matrix[0].length;
int lo = 0, hi = rowNum * colNum - 1;
while(lo <= hi) {
int mid = lo + (hi - lo) / 2;
int row = mid / colNum, col = mid % colNum;
if(matrix[row][col] < target)
lo = mid + 1;
else if(matrix[row][col] > target)
hi = mid - 1;
else
return true;
}
return false;
}
}
二刷：
跟一刷一样，利用二分搜索， 然后在搜索的时候把mid转化为矩阵的行和列坐标就可以了。
Time Complexity - O(logmn)， Space Complexity - O(1).
Java:
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0) {
return false;
}
int rowNum = matrix.length, colNum = matrix[0].length;
int lo = 0, hi = rowNum * colNum - 1;
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
int row = mid / colNum;
int col = mid % colNum;
if (matrix[row][col] == target) {
return true;
} else if (matrix[row][col] < target) {
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return false;
}
}
三刷:
把2D Matrix转换为1D Array，然后使用Binary Search就可以了。假设1D Array中的index是mid，转换就是 row = mid / colNum， col = mid % colNum
Java:
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0) { return false; }
int rowNum = matrix.length, colNum = matrix[0].length;
int lo = 0, hi = rowNum * colNum - 1;
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
int rowMid = mid / colNum;
int colMid = mid % colNum;
if (matrix[rowMid][colMid] == target) { return true; }
else if (matrix[rowMid][colMid] < target) { lo = mid + 1; }
else { hi = mid - 1; }
}
return false;
}
}
测试：