Search a 2D Matrix leetcode

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.

先对矩阵的行数进行二分法查找， 找到target所在行数 再对此行二分法查找 来判断target时候存在。

这个的算法时间复杂度是O(log(rows)+log(columns))。
做这道题时候犯二了 在第一次二分法的时候先判断matrix[start][0] < target结果死活不对， 后来发现如果这么判定的话当matrix[end][0] < target 时候也会跳到start行。

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0){
            return false;
        }
        if (matrix[0] == null || matrix[0].length == 0){
            return false;
        }
        int start = 0;
        int end = matrix.length - 1;
        int row;
        while (start + 1 < end){// find the row
            int mid = start + (end - start) / 2;
            if (matrix[mid][0] == target){
                return true;
            } else if (matrix[mid][0] < target){
                start = mid;
            } else {
                end = mid;
            }
        }
        if (matrix[end][0] <= target) {
            row = end;
        } else if (matrix[start][0] <= target) {
            row = start;
        } else {
            return false;
        }
        start = 0;
        end = matrix[0].length - 1;
        while (start + 1 < end){// find the column
            int mid = start + (end - start) / 2;
            if (matrix[row][mid] == target){
                return true;
            } else if (matrix[row][mid] < target){
                start = mid;
            } else {
                end = mid;
            }
        }
        if (matrix[row][start] == target){
            return true;
        } else if (matrix[row][end] == target){
            return true;
        } else {
            return false;
        }
    }
}

第二种方法
把2d矩阵转换成1d 赋值start = 0 end = row * column -1
每个元素都可以用 matrix[position/column][position%column]来表示
然后用2分法解题
时间复杂度 O(log(row * column))

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0){
            return false;
        }
        if (matrix[0] == null || matrix[0].length == 0){
            return false;
        }
        int row = matrix.length;
        int column = matrix[0].length;
        int start = 0;
        int end = row * column - 1;
        while (start + 1 < end){
            int mid = start + (end - start) / 2;
            int num = matrix[mid / column][mid % column];
            if (num == target){
                return true;
            } else if(num < target){
                start = mid;
            } else {
                end = mid;
            }
        }
        if (matrix[start / column][start % column] == target){
            return true;
        } else if (matrix[end / column][end % column] == target){
            return true;
        } else {
            return false;
        }
    }
}

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0) {
            return false;
        }
        if (matrix[0] == null || matrix[0].length == 0) {
            return false;
        }
        int m = matrix.length;
        int n = matrix[0].length;
        int total = m * n;
        int start = 0;
        int end = total - 1;
        while (start <= end) {
            int mid = start + (end - start) / 2;
            if (matrix[mid/n][mid%n] == target) {
                return true;
            }
            else if (matrix[mid/n][mid%n] > target) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }
        return false;
    }
}