Write an efficient algorithm that searches for a value in an m x n matrix, return the occurrence of it.

This matrix has the following properties:

* Integers in each row are sorted from left to right.

* Integers in each column are sorted from up to bottom.

* No duplicate integers in each row or column.

Example

Consider the following matrix:

[

    [1, 3, 5, 7],

    [2, 4, 7, 8],

    [3, 5, 9, 10]

]

Given target = 3, return 2.

Challenge

O(m+n) time and O(1) extra space

Solution:

 public class Solution {
/**
* @param matrix: A list of lists of integers
* @param: A number you want to search in the matrix
* @return: An integer indicate the occurrence of target in the given matrix
*/
public int searchMatrix(ArrayList<ArrayList<Integer>> matrix, int target) {
int m = matrix.size();
if (m==0) return 0;
int n = matrix.get(0).size();
if (n==0) return 0; return searchMatrixRecur(matrix,target,0,0,m-1,n-1);
} public int searchMatrixRecur(ArrayList<ArrayList<Integer>> matrix, int target, int x1, int y1, int x2, int y2){
if (x2<x1 || y2<y1) return 0; if (x1==x2 && y1==y2)
if (matrix.get(x1).get(y1)==target) return 1;
else return 0; int midX = (x1+x2)/2;
int midY = (y1+y2)/2;
int midVal = matrix.get(midX).get(midY);
int res = 0; if (midVal==target){
//We have to search all the four sub matrix.
res++;
res += searchMatrixRecur(matrix,target,x1,y1,midX-1,midY-1);
res += searchMatrixRecur(matrix,target,midX+1,midY+1,x2,y2);
res += searchMatrixRecur(matrix,target,(x1+x2)/2+1,y1,x2,(y1+y2)/2-1);
res += searchMatrixRecur(matrix,target,x1,(y1+y2)/2+1,(x1+x2)/2-1,y2);
} else if (midVal>target) {
int leftX = (x1+x2)/2;
int leftY = y1;
int upX = x1;
int upY = (y1+y2)/2;
if (target==matrix.get(leftX).get(leftY)) res++;
if (target==matrix.get(upX).get(upY)) res++;
if (target <= matrix.get(leftX).get(leftY) && target <=matrix.get(upX).get(upY)){
res += searchMatrixRecur(matrix,target,x1,y1,midX-1,midY-1);
} else if (target <= matrix.get(leftX).get(leftY)){
res += searchMatrixRecur(matrix,target,x1,y1,(x1+x2)/2-1,y2);
} else if (target <= matrix.get(upX).get(upY)){
res += searchMatrixRecur(matrix,target,x1,y1,x2,(y1+y2)/2-1);
} else {
res += searchMatrixRecur(matrix,target,x1,y1,x2,(y1+y2)/2-1);
res += searchMatrixRecur(matrix,target,upX,upY,(x1+x2)/2-1,y2);
}
} else {
int rightX = (x1+x2)/2;
int rightY = y2;
int lowX = x2;
int lowY = (y1+y2)/2;
if (target==matrix.get(rightX).get(rightY)) res++;
if (target==matrix.get(lowX).get(lowY)) res++;
if (target >= matrix.get(rightX).get(rightY) && target >= matrix.get(lowX).get(lowY)){
res += searchMatrixRecur(matrix,target,midX+1,midY+1,x2,y2);
} else if (target >= matrix.get(rightX).get(rightY)){
res += searchMatrixRecur(matrix,target, (x1+x2)/2+1,y1,x2,y2);
} else if (target >= matrix.get(lowX).get(lowY)){
res += searchMatrixRecur(matrix,target, x1, (y1+y2)/2+1, x2, y2);
} else {
res += searchMatrixRecur(matrix,target, (x1+x2)/2+1,y1, lowX, lowY);
res += searchMatrixRecur(matrix,target, x1, (y1+y2)/2+1, x2, y2);
} }
return res;
}
}

LintCode-Search 2D Matrix II的更多相关文章

  1. LintCode 38. Search a 2D Matrix II

    Write an efficient algorithm that searches for a value in an m x n matrix, return the occurrence of ...

  2. Search a 2D Matrix | & II

    Search a 2D Matrix II Write an efficient algorithm that searches for a value in an m x n matrix, ret ...

  3. leetcode 74. Search a 2D Matrix 、240. Search a 2D Matrix II

    74. Search a 2D Matrix 整个二维数组是有序排列的,可以把这个想象成一个有序的一维数组,然后用二分找中间值就好了. 这个时候需要将全部的长度转换为相应的坐标,/col获得x坐标,% ...

  4. 【LeetCode】240. Search a 2D Matrix II

    Search a 2D Matrix II Write an efficient algorithm that searches for a value in an m x n matrix. Thi ...

  5. LeetCode -- Search a 2D Matrix & Search a 2D Matrix II

    Question: Search a 2D Matrix Write an efficient algorithm that searches for a value in an m x n matr ...

  6. 240.Search in a 2D Matrix II

    /* * 240.Search in a 2D Matrix II * 2016-6-17by Mingyang * From left-bottom to right-top * 他这道题目虽说是用 ...

  7. Leetcode之二分法专题-240. 搜索二维矩阵 II(Search a 2D Matrix II)

    Leetcode之二分法专题-240. 搜索二维矩阵 II(Search a 2D Matrix II) 编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target.该矩阵 ...

  8. LeetCode 240. 搜索二维矩阵 II(Search a 2D Matrix II) 37

    240. 搜索二维矩阵 II 240. Search a 2D Matrix II 题目描述 编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target.该矩阵具有以下特性 ...

  9. 【刷题-LeetCode】240. Search a 2D Matrix II

    Search a 2D Matrix II Write an efficient algorithm that searches for a value in an m x n matrix. Thi ...

  10. 【Lintcode】038.Search a 2D Matrix II

    题目: Write an efficient algorithm that searches for a value in an m x n matrix, return the occurrence ...

随机推荐

  1. Set集合——HashSet、TreeSet、LinkedHashSet(2015年07月06日)

    一.Set集合不同于List的是: Set不允许重复 Set是无序集合 Set没有下标索引,所以对Set的遍历要通过迭代器Iterator 二.HashSet 1.HashSet由一个哈希表支持,内部 ...

  2. JS测试浏览器类型的代码

    function getOs(url,title) { var OsObject = ""; if(navigator.userAgent.indexOf("MSIE&q ...

  3. .NET DLL 保护措施详解(四)各操作系统运行情况

    我准备了WEB应用程序及WinForm应用程序,分别在WIN SERVER 2012/2008/2003.Win7/10上实测,以下为实测结果截图: 2012 2008 2003 WIN7 WIN10 ...

  4. c#中文件上传(1)

    * * ;//3M picPath = Server.MapPath("........."); HttpFileCollection postfile = Context.Req ...

  5. sql语句将本地服务器中的数据插入到外网服务器中

    --将本地的数据库中的某张表中的数据导入到180的数据库中 --这个要在本地的数据库运行 exec sp_addlinkedserver 'srv_lnk', '', 'SQLOLEDB','xxx. ...

  6. java中serializable

    java中serializable是一个对象序列化的接口,一个类只有实现了Serializable接口,它的对象才是可序列化的.因此如果要序列化某些类的对象,这些类就必须实现Serializable接 ...

  7. OpenStack 控制台不能不能访问的问题

    经过一个多月断断续续的OpenStack部署,今天终于搞定基本的云环境部署,linux.Windows虚拟机都可以正常运行!虽然期间遇到了N多的坑,在自己不断学习,不断找E文的过程中都逐一被我攻破,但 ...

  8. 30类css选择器

    大概大家都知道id,class以及descendant选择器,并且整体都在使用它们,那么你正在错误拥有更大级别的灵活性的选择方式.这篇文章里面提到的大部分选择器都是在CSS3标准下的,所以它们只能在相 ...

  9. score

    #include<iostream> using namespace std; class student{ public: int Input() { ;i<;i++) { cou ...

  10. Ubuntu系统下允许Apache的mod_rewrite功能

    首先,使能apache的rewirte模块,在shell里输入下边的命令: sudo a2enmod rewrite 然后重启一下webserver使更改生效 sudo service apache2 ...