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：