LeetCode刷题【Array】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.

解决方法一：Runtime: 1 ms 二分查找思想

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

解决方法二： 把二维数组当做Array处理，可能存在m*nRuntime: 1 ms

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

参考：

【1】https://leetcode.com/