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 sort ...
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其他好文 时间:
2017-05-29 11:57:06
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154
题目: Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place. 链接: http://leetcode.com/problems/set-matrix-zeroes/ ...
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其他好文 时间:
2017-05-29 11:54:35
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172
https://leetcode.com/problems/01-matrix/ 思路参考:http://www.cnblogs.com/grandyang/p/6602288.html ...
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其他好文 时间:
2017-05-29 10:02:32
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190
Problem statement: Given an m * n matrix M initialized with all 0's and several update operations. Operations are represented by a 2D array, and each ...
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其他好文 时间:
2017-05-29 09:59:23
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154
传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=4128 【题解】 矩阵版本的BSGS。 至于如何不需要求逆,详见:http://www.cnblogs.com/galaxies/p/bzoj2480.html # include <map> ...
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其他好文 时间:
2017-05-28 15:29:32
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217
In MATLAB, there is a very useful function called 'reshape', which can reshape a matrix into a new one with different size but keep its original data. ...
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其他好文 时间:
2017-05-28 14:58:12
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191
Given an m * n matrix M initialized with all 0's and several update operations. Operations are represented by a 2D array, and each operation is repres ...
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其他好文 时间:
2017-05-28 11:51:50
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168
CGAffineTransform matrix = CGAffineTransformMake(1, 0, tanf(15 * (CGFloat)M_PI / 180), 1, 0, 0); UIFontDescriptor *desc = [ UIFontDescriptor fontDescr ...
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移动开发 时间:
2017-05-28 10:04:45
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402
输入:只包含0,1的矩阵 输出:元素1到达最近0的距离 算法思想:广度优先搜索。 元素为0为可达区域,元素为1为不可达区域,我们的目标是为了从可达区域不断地扩展至不可达区域,在扩展的过程中,也就计算出了这些不可达区域到达最近可达区域的距离。 每个可达元素都记录了到当前位置的距离,因此在后续的遍历中, ...
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其他好文 时间:
2017-05-27 23:01:53
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206
#include <cstdio> #include <cstring> #include <algorithm> #include <string> #include <cmath> #include <iostream> using namespace std; bool fun(double ...
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其他好文 时间:
2017-05-27 14:25:15
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143