leetcode------Search a 2D Matrix

标题：	Search a 2D Matrix
通过率	31.3%
难度	中等

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.

本题思路清晰，就是再一个二维数组中找某个元素，那么立即想到了二分法，关键点，按行算，第k个元素的行数等于k/列数，列数等于k%列数。

还有一种思路就是，每一行的开头一定比上一行全部元素都大，那么从最后一行的第一个开始，每次比较每行的第一个，找到第一个比target小的元素，也就说明了target若存在一定在那一行，

1、二分法查找代码如下：

 1 public class Solution {

 2     public boolean searchMatrix(int[][] matrix, int target) {

 3         if(matrix==null || matrix.length==0 || matrix[0].length==0) 

 4             return false;

 5  

 6         int m = matrix.length;

 7         int n = matrix[0].length;

 8  

 9         int start = 0;

10         int end = m*n-1;

11  

12         while(start<=end){

13             int mid=(start+end)/2;

14             int midX=mid/n;

15             int midY=mid%n;

16  

17             if(matrix[midX][midY]==target) 

18                 return true;

19  

20             if(matrix[midX][midY]<target){

21                 start=mid+1;

22             }else{

23                 end=mid-1;

24             }

25         }

26  

27         return false;

28     }

29 }

2、查找第一个小的数：

 1 public class Solution {

 2     public boolean searchMatrix(int[][] matrix, int target) {

 3         int m=matrix.length;

 4         int n=matrix[0].length;

 5         int i=m-1;

 6         int j=0;

 7         while(i>=0){

 8             if(matrix[i][0]>target)i--;

 9             else break;

10         }

11         if(i<0)return false;

12         while(j<n){

13             if(matrix[i][j]==target)return true;

14             else j++;

15         }

16         return false;

17         

18     }

19 }

3、python代码：

 1 class Solution:

 2     def searchMatrix(self, matrix, target):

 3         i, j = 0, len(matrix[0]) - 1

 4         while i < len(matrix) and j >= 0:

 5             if target < matrix[i][j]:

 6                 j -= 1

 7             elif target > matrix[i][j]:

 8                 i += 1

 9             else:

10                 return True

11         return False