[Yahoo] Cloest palindrome number, Solution

本文介绍了一种算法,用于找到给定整数最接近的回文数。通过将数字转换为字符串并操作其一半来创建可能的回文,然后计算这些回文与原始数字之间的绝对差值以确定最接近的回文数。

Given an integer, print the closest number to it that is a palindrome – eg, the number “1224” would return “1221”.

[Thoughts]

pseudo code: (with two examples in parentheses)





- Convert the number into string. (1224, 39999)


- take half of the string. ( "12", "399" )


- copy first half to second half in reverse order (take care of no of chars)
( "12" -> "1221", "399" -> "39993" )


- convert to number and measure the abs. difference with original number - diff1
( |1221 - 1224| = 3, |39993-39999| = 6)


- add 1 to half string and now copy first half to second half in reverse order
( 12+1 = 13, 399 + 1 = 400, 13-> 1331, 400->40004)


- convert to number and measure the abs. difference with original number - diff2
( |1331-1224| = 107, |40004-39999| = 5 )


- if diff1<diff2 return first number else return second number
( 1221, 40004)
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function [move_points,T] = SparsePointToPoint(source_points,target_points,max_icp,max_outer,max_inner,p) %SPARSE_POINT_TO_POINT sparse point to point % INPUT: % source_points: source point clouds % target_points: target point clouds % max_icp: maximum iteration number of searching cloest points and % solving R and t % max_out: maximum iteration number of solving ALM by ADMM % max_in: maximum iteration number of solving R and t % p: p-norm % OUTPUT: % move_points: final results % T: optimal homogenous transformation matrix fprintf('sparse point-to-point:\n'); if length(source_points(:,1))==4 source_points=source_points(1:3,:); target_points=target_points(1:3,:); end move_points = source_points; NS = createns(target_points','NSMethod','kdtree'); Num=length(source_points(1,:)); old_points=source_points; lambda=zeros(size(move_points)); Z=zeros(size(move_points)); mu=10; T=eye(4); for icp=1:max_icp [idx, ~] = knnsearch(NS,move_points','k',1); match_points= target_points(:,idx); fprintf('iteration at %d-%d\n', icp,max_icp); for i=1:max_outer for j=1:max_inner H=move_points-match_points+lambda/mu; for k=1:Num Z(:,k)=shrink(H(:,k),p,mu); end C=match_points+Z-lambda / mu; [R,t]=svd_icp(move_points,C); T1=eye(4); T1(1:3,1:3)=R; T1(1:3,4)=t; T=T1*T; move_points=R*move_points+t; dual_max=0; for m=1:Num dual=norm((old_points(:,m)-move_points(:,m))); if (dual>dual_max) dual_max=dual; end end old_points=move_points; if dual_max <1e-5 break; end end delta=move_points-match_points-Z; prime_max=0; for k=1:Num prime=norm(move_points(:,k)-match_points(:,k)-Z(:,k)); if (prime>prime_max) prime_max=prime; end end lambda=lambda+mu*delta; if dual_max <1e-5 && prime_max<1e-5 break; end end end end function [R,t]=svd_icp(source_points,target_points) centerA=mean(source_points')'; centerB=mean(target_points')'; tempA=source_points-centerA; tempB=target_points-centerB; H=zeros(3,3); for i=1:length(source_points(1,:)) H=H+tempA(:,i)*tempB(:,i)'; end [U,~,V]=svd(H); R=V*U'; t=centerB-R*centerA; end 使用PCL重写该matlab代码
07-02
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