原题地址:https://oj.leetcode.com/problems/anagrams/题意:Given
an array of strings, return all groups of strings that are anagrams.Note: All
inputs will be i...
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编程语言 时间:
2014-06-29 13:25:28
阅读次数:
355
用减法可能会超时,但可以用二分
class Solution {
public:
int divide(int d1, int d2) {// d1/d2
if(d1==0)
return 0;
if(d2==1)
return d1;
if(d2==-1)
...
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其他好文 时间:
2014-06-29 07:22:02
阅读次数:
208
归并排序是建立在归并操作上的一种有效的排序算法。该算法是採用分治法(Divide and
Conquer)的一个很典型的应用。首先考虑下怎样将将二个有序数列合并。这个很easy,仅仅要从比較二个数列的第一个数,谁小就先取谁,取了后就在相应数列中删除这个数。然后再进行比較,假设有数列为空,那直接将还有...
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其他好文 时间:
2014-06-11 21:52:36
阅读次数:
289
Divide two integers without using
multiplication, division and mod operator.
其实刚开始看到这道题的时候,感觉应该是略简单。但真正开始写的时候发现了很多错误。 最开始的想法就是divisor一个一个加上去直到大于divide...
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其他好文 时间:
2014-06-11 12:17:22
阅读次数:
274
Grouping Records Together Based on a Field
Problem
You
have a sequence of dictionaries or instances and you want to iterate over the data in groups based on the value of a...
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编程语言 时间:
2014-06-08 18:22:29
阅读次数:
280
package chap04_Divide_And_Conquer;import static
org.junit.Assert.*;import java.util.Arrays;import org.junit.Test;/** * 算反导论第四章
4.1 最大子数组 * * @author ....
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其他好文 时间:
2014-06-07 20:21:38
阅读次数:
212
Let d(n) be defined as the sum of proper divisors of n (numbers less than
n which divide evenly into n).
If d(a) = b and d(b) = a, where a
b, then a and b are an amicable pair and each of a and
b...
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其他好文 时间:
2014-06-07 13:40:53
阅读次数:
226
Divide two integers without using
multiplication, division and mod operator. 1 class Solution { 2 public: 3 int
divide(int dividend, int divisor) ...
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其他好文 时间:
2014-06-06 17:40:07
阅读次数:
193
Table Groups [AX 2012]0 out of 1 rated this
helpful-Rate this topic Updated:February 21, 2012 Applies To:Microsoft Dynamics
AX 2012 R2, Microsoft Dyna...
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其他好文 时间:
2014-06-05 20:45:09
阅读次数:
326
问题描述:
Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a b, then a and b are
an amicable pair and each ...
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编程语言 时间:
2014-06-02 12:31:42
阅读次数:
298