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. 水题2枚 Codevs1464&&Codevs1472

    1472 体检  时间限制: 1 s  空间限制: 64000 KB  题目等级 : 白银 Silver 题解  查看运行结果     题目描述 Description 郑厂长不是正厂长 也不是副厂长 ...

  2. 【.NET基础】--委托、事件、线程(2)

    本文介绍event的使用以及原理,本文接上一篇文章的Demo继续[下载上一篇Demo] 上一篇我们在类(dg_SayHi.cs)里面定义代理了4个Delegate,然后在Button的后台事件中 新建 ...

  3. sql语句使用游标修改表中数据

    declare @a varchar(),@b varchar() declare user_cursor cursor for select a,b from tableA tab open use ...

  4. C#_字符串的操作

    1: using System; 2: using System.Collections.Generic; 3: using System.Linq; 4: using System.Text; 5: ...

  5. 【网络收集】获取JavaScript 的时间使用内置的Date函数完成

    var mydate = new Date(); mydate.getYear(); //获取当前年份(2位) mydate.getFullYear(); //获取完整的年份(4位,1970-???? ...

  6. JavaScript之canvas

    num.push(x,y); 动画草图(举个栗子,我们把数字“2”给画出来): <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transit ...

  7. Javascript之数据执行原理探究

    Javascript在Web服务器端执行原理: 1.客户端请求数据,即我们在上网时在地址栏中输入某个网址,浏览器接收到数据之后,向远程web服务器发送请求报文. 2.web服务器响应请求,web服务器 ...

  8. CSS里的引用@import、link

    引入CSS的方法有两种,一种是@import,一种是link @import url('地址');<link href="地址" rel="stylesheet&q ...

  9. UIView的frame的扩展分类,轻松取出x、y、height、width等值

    一.引言: 在ios开发中,就界面搭建.控件布局时,都会很恶心的通过很长的代码才能取出控件的x.y.height.width等值,大大降低了开发效率.那为了省略这些恶心的步骤,小编在这里给UIView ...

  10. OC4_遵守多个协议

    // // Calulator.h // OC4_遵守多个协议 // // Created by zhangxueming on 15/6/24. // Copyright (c) 2015年 zha ...