lintcode28- Search a 2D Matrix- easy

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.

Example

Consider the following matrix:

[
    [1, 3, 5, 7],
    [10, 11, 16, 20],
    [23, 30, 34, 50]
]

Given target = 3, return true.

Challange

O(log(n) + log(m)) time

---

1. 两次binary search

先针对每行第一个元素检索，找到可能行；再在可能行内检索。

public class Solution {
    /*
     * @param matrix: matrix, a list of lists of integers
     * @param target: An integer
     * @return: a boolean, indicate whether matrix contains target
     */
    public boolean searchMatrix(int[][] matrix, int target) {

        //看看这样写行不
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0){
            return false;
        }

        // 对吗
        int rows = matrix.length;
        int cols = matrix[0].length;

        //find the possible row.
        int start = 0;
        int end = rows - 1;
        int possibleRow = 0;
        while (start + 1 < end){
            int mid = start + (end - start) / 2;
            if (target < matrix[mid][0]){
                end = mid;
            } else if (target == matrix[mid][0]){
                return true;
            } else {
                start = mid;
            }
        }
        if (matrix[end][0] > target){
            possibleRow = start;
        } else if (matrix[end][0] == target){
            return true;
        } else {
            possibleRow = end;
        }

        //find in the possible row.
        start = 0;
        cols = matrix[possibleRow].length;
        end = cols - 1;
        while (start + 1 < end){
            int mid = start + (end - start) / 2;
            if (target < matrix[possibleRow][mid]){
                end = mid;
            } else if (target == matrix[possibleRow][mid]){
                return true;
            } else {
                start = mid;
            }
        }
        if (matrix[possibleRow][start] == target || matrix[possibleRow][end] == target){
            return true;
        }
        return false;
    }
}

---

2.一次binary search

把所有数字连起来看做一个排序一元数组，用/和%提取行列坐标，一次binary search。

public class Solution {
    /*
     * @param matrix: matrix, a list of lists of integers
     * @param target: An integer
     * @return: a boolean, indicate whether matrix contains target
     */
    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 rows = matrix.length;
        int cols = matrix[0].length;

        int start = 0;
        int end = rows * cols - 1;

        while (start + 1 < end){
            int mid = start + (end - start) / 2;
            if (target < matrix[mid / cols][mid % cols]){
                end = mid;
            } else if (target == matrix[mid / cols][mid % cols]){
                return true;
            } else {
                start = mid;
            }
        }
        if (matrix[start / cols][start % cols] == target || matrix[end / cols][end % cols] == target){
            return true;
        }
        return false;
    }
}

---

1.对二元数组的输入判别：

if(matrix == null || matrix.length == 0){
            return false;
}
        
if(matrix[0] == null || matrix[0].length == 0){
            return false;
}

2.对二元数组的行列长度提取：（要在判别存在行以后才可以用matrix[0]）

int row = matrix.length;
int column = matrix[0].length;