CC150 - 11.6
Question:
Given an M x N matrix in which each row and each column is sorted in ascending order, write a method to find an element.
package POJ;
public class Main {
/**
*
* 11.6 Given an M x N matrix in which each row and each column is sorted in
* ascending order, write a method to find an element.
*
*/
public static void main(String[] args) {
Main so = new Main();
}
public static Coordinate findElement(int[][] matrix, int x) {
Coordinate origin = new Coordinate(0, 0);
Coordinate dest = new Coordinate(matrix.length - 1,
matrix[0].length - 1);
return findElement(matrix, origin, dest, x);
}
public static Coordinate findElement(int[][] matrix, Coordinate origin,
Coordinate dest, int x) {
if (!origin.inbounds(matrix) || !dest.inbounds(matrix)) {
return null;
}
if (matrix[origin.row][origin.column] == x)
return origin;
if (!origin.isBefore(dest))
return null;
Coordinate start = (Coordinate) origin.clone();
int diagDist = Math.min(dest.row - origin.row, dest.column
- origin.column);
Coordinate end = new Coordinate(start.row + diagDist, start.column
+ diagDist);
Coordinate p = new Coordinate(0, 0);
while (start.isBefore(end)) {
p.setToAverage(start, end);
if (x > matrix[p.row][p.column]) {
start.row = p.row + 1;
start.column = p.column + 1;
} else {
end.row = p.row - 1;
end.column = p.column - 1;
}
}
return partitionAndSearch(matrix, origin, dest, start, x);
}
private static Coordinate partitionAndSearch(int[][] matrix,
Coordinate origin, Coordinate dest, Coordinate pivot, int elem) {
// TODO Auto-generated method stub
Coordinate lowerLeftOrigin = new Coordinate(pivot.row, origin.column);
Coordinate lowerLeftDest = new Coordinate(dest.row, pivot.column - 1);
Coordinate upperRightOrigin = new Coordinate(origin.row, pivot.column);
Coordinate upperRightDest = new Coordinate(pivot.row - 1, dest.column);
Coordinate lowerLeft = findElement(matrix, lowerLeftOrigin,
lowerLeftDest, elem);
if (lowerLeft == null)
return findElement(matrix, upperRightOrigin, upperRightDest, elem);
return lowerLeft;
}
}
class Coordinate implements Cloneable {
int row;
int column;
public Coordinate(int r, int c) {
row = r;
column = c;
}
public boolean inbounds(int[][] matrix) {
return row >= 0 && column >= 0 && row < matrix.length
&& column < matrix[0].length;
}
public boolean isBefore(Coordinate p) {
return row <= p.row && column <= p.column;
}
public Object clone() {
return new Coordinate(row, column);
}
public void setToAverage(Coordinate min, Coordinate max) {
row = (min.row + max.row) / 2;
column = (min.column + max.column) / 2;
}
}
CC150 - 11.6的更多相关文章
- CC150 - 11.5
Question: Given a sorted array of strings which is interspersed with empty strings, write a method t ...
- CC150 - 11.3
Question: Given a sorted array of n integers that has been rotated an unknown number of times, write ...
- CC150 - 11.2
Question: Write a method to sort an array of strings so that all the anagrams are next to each other ...
- CC150 - 11.1
Question: You are given two sorted arrays, A and B, where A has a large enough buffer at the end to ...
- 地区sql
/*Navicat MySQL Data Transfer Source Server : localhostSource Server Version : 50136Source Host : lo ...
- 11.8---维护x的秩(CC150)
思路:比较easy.就是借助hashset让他有序然后就能够比较节省时间了. 答案: public static int[] getRankOfNumber(int[] a, int n){ int[ ...
- 11.7---叠罗汉表演节目(CC150)
1,牛客网第一题:这其实跟找最长递增子序列是一个东西.注意的地方是,返回的是最大的dp,而不是dp[N-1]. 答案: public static int getHeight(int[] men, i ...
- 11.6---矩阵查找元素(CC150)
思路,一旦提到查找就要想到二分查找. public static int[] findElement(int[][] a, int n, int m, int key) { // write code ...
- 11.5---含有空字符串的字符串查找(CC150)
注意,1,"" 和 " ".是不同的,空字符串指的是"": 2,注意String的compareTo.小于是指<0.并不是==-1: ...
随机推荐
- CentOS6.5以runlevel 3开机时自动连接某无线设置示例
[参考]http://blog.csdn.net/simeone18/article/details/8580592 [方法一] 假设无线网卡代号为wlan0,无线AP的essid为:TheWiFi, ...
- iOS UIDatePicker frame改变问题
这种方法不行: pickerCtl = UIDatePicker(frame:pickerFrame) 但是这种却行 pickerCtl = UIDatePicker() pickerCtl!.fra ...
- web页面版权部分的显示问题
网站开发中绝大部分页面底部都需要版权信息,一般都是Copyright ©域名 2014 - 2015. All Rights Reserved.这种格式,当然也有其他的,有时候不太注意会发现做出的这个 ...
- 七维互联(www.7wei.com)
七维互联 http://www.7wei.com/ 黄云贵的Delphi http://www.cnblogs.com/huangygdelphi/articles/2232171.htm ...
- 如何为Linux生成和打上patch
通过diff工具生成补丁, patch工具打上补丁. 在使用diff之前, 你需要保留一份未修改过的源码, 然后在其它地方修改源码的一份拷贝. diff对比这两份源码生成patch. 修改过的源码必须 ...
- SQL 查询CET使用领悟
用到sql的遍历循环查询,如果不考虑用CET,估计又到了自己造轮子的时代了,现在觉得sql的CET确实是个好东西,针对SQL的递归查询,很是不错的方法: with etcRecommandINfo2( ...
- hdu 1241:Oil Deposits(DFS)
Oil Deposits Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total ...
- 免费电子书:使用VS Online敏捷管理开源项目
今天推荐的是一本由微软出版社发布的免费电子书,涉及的话题是如何在Visual Studio Online中基于敏捷的思想来管理开源项目. 本书的几位作者(自称ALM领域的游侠),给大家分享了在一个敏捷 ...
- loj 1108(spfa判负环)
题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=26823 思路:题目的意思是求出所有的能够到达负环的点.负环很好求, ...
- PHP API 框架开发的学习
基于互联网的应用正变得越来越普及,在这个过程中,有更多的站点将自身的资源开放给开发者来调用.对外提供的API 调用使得站点之间的内容关联性更强,同时这些开放的平台也为用户.开发者和中小网站带来了更大的 ...