[LintCode] 28 Search a 2D Matrix 解题报告

Description
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



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



思路
二分查找。
把二维数组看成一维数组,然后进行二分查找。
一维数组的index对应二维数组的[index / n][index % n]。
n是matrix[0].length。


Code

public class Solution {
    /**
     * @param 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) {
        // write your code here
        
        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 start = 0;
        int end = m * n - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (matrix[mid / n][mid % n] < target) {
                start = mid;
            }
            else {
                end = mid;
            }
        }
        
        if (matrix[start / n][start % n] == target) {
            return true;
        }
        if (matrix[end / n][end % n] == target) {
            return true;
        }
        return false;
    }
}