MS Brings Cortana Enabled Cards On Bing.com - 未名空间MITBBS历史存档

国际科技财经博客移民网络热点娱乐民生时事公众号

Redian新闻

>未名空间

>PDA - 掌中宝

MS Brings Cortana Enabled Cards On Bing.com

MS Brings Cortana Enabled Cards On Bing.com# PDA - 掌中宝

e*e2014-04-16 07:04

1 楼

Given a vector of integers, find the longest consecutive sub-sequence of
increasing numbers. If two sub-sequences have the same length, use the one
that occurs first. An increasing sub-sequence must have a length of 2 or
greater to qualify.
Example input:
[1 0 1 2 3 0 4 5]
Result:
[0 1 2 3]

s*m2014-04-16 07:04

2 楼

Microsoft is bringing this personalized experience to the web starting today
. Microsoft is unveiling a set of personalized cards on the Bing homepage
that will help you keep track of things that matter to you. Set up your
interests in Bing settings and you will start seeing personalized news,
weather, flights, and stocks with more to come – all as part of the search
experience.

e*e2014-04-16 07:04

3 楼

我的解法，总觉得用Java写还有更好的。
public class LongestConsecutiveSubsequence {

private int longestLCSStartPos = 0;
private int longestLCSCount = 0;
public Vector lcs(Vector input) {
if( input == null || input.size() < 2 )
throw new RuntimeException( "Invalid input." );

lcs( input, 0 );
if ( longestLCSCount < 2 )
throw new RuntimeException( "lcs length is less than 2." );

Vector r = new Vector();
for( int i = 0 ; i < longestLCSCount; i++ )
r.add( input.get( i + longestLCSStartPos ) );
return r;
}

private void lcs(Vector input, int startPos ) {
if( startPos == input.size() - 1 )
return;

int lcsStartPos = findNextBottomPos( input, startPos );
int lcsEndPos = findNextTopPos( input, lcsStartPos );
int lcsCount = lcsEndPos - lcsStartPos + 1;

if( lcsCount > longestLCSCount) {
longestLCSStartPos = lcsStartPos;
longestLCSCount = lcsCount;
}

lcs( input, lcsEndPos );
}

private int findNextBottomPos( Vector input, int startPos ) {
while( startPos + 1 < input.size() && input.get( startPos + 1 ) <=
input.get( startPos ) )
startPos++;
return startPos;
}

private int findNextTopPos( Vector input, int startPos ) {
while( startPos + 1 < input.size() && input.get( startPos + 1 ) >
input.get( startPos ) )
startPos++;
return startPos;
}
}

r*h2014-04-16 07:04

4 楼

连续递增子串的话，从左往右扫一轮不就可以了吗
int i=0, j, maxLength=0;
while(i < A.length-1){
while(A[i]> A[i+1] && ii++;
j = i+1;
while(jA[j-1])
j++;
if(maxLength < j-i)
maxLength = j-i;
i = j;
}
if(maxLength <= 1)
return 0;
return maxLength;

【在 e****e 的大作中提到】

: Given a vector of integers, find the longest consecutive sub-sequence of
: increasing numbers. If two sub-sequences have the same length, use the one
: that occurs first. An increasing sub-sequence must have a length of 2 or
: greater to qualify.
: Example input:
: [1 0 1 2 3 0 4 5]
: Result:
: [0 1 2 3]