74. Search a 2D Matrix

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：