Algorithm Problems
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Revision as of 03:40, 28 July 2014 by Hall (Talk | contribs) (→Calculate degree between hour and minute hand)
Contents
Binary Tree
All about Binary search tree
All about Linked List
Print linklist inversed
/////////////////////////////////////////////////////////////////////// // Print a list from end to beginning // Input: pListHead - the head of list /////////////////////////////////////////////////////////////////////// void PrintListReversely(ListNode* pListHead) { if(pListHead != NULL) { // Print the next node first if (pListHead->m_pNext != NULL) { PrintListReversely(pListHead->m_pNext); } // Print this node printf("%d", pListHead->m_nKey); } }
Inverse Array
for (int i = 0; i < b/2; i=i++){ char temp1 = a[i]; a[i] = a[(b-1)-i]; a[(b-1)-i] = temp1; cout << a[i] ; }
Find duplicate numbers in array
<?php /* Find duplicate number in array */ //construct $b[n]=n $b=array(); $duplicateArr=array(); $a=array("1","3","4","1","5","2","8","3","2244","1"); for($i=0;$i<sizeof($a);$i++){ if(isset($b[$a[$i]])) array_push($duplicateArr,$a[$i]); else $b[$a[$i]]=$a[$i]; } var_dump($duplicateArr); ?>
Interpolation search
public int interpolationSearch(int[] sortedArray, int toFind){ // Returns index of toFind in sortedArray, or -1 if not found int low = 0; int high = sortedArray.length - 1; int mid; while (sortedArray[low] <= toFind && sortedArray[high] >= toFind) { mid = low + ((toFind - sortedArray[low]) * (high - low)) / (sortedArray[high] - sortedArray[low]); //out of range is possible here if (sortedArray[mid] < toFind) low = mid + 1; else if (sortedArray[mid] > toFind) // Repetition of the comparison code is forced by syntax limitations. high = mid - 1; else return mid; } if (sortedArray[low] == toFind) return low; else return -1; // Not found }
Random Selector
Select k smallest value in array. O(n)
<?php function swap(&$a,&$b){ if($a!=$b) $a=$a+$b-($b=$a); //$a^=$b^=$a^=$b; } function partition(&$listArr, $left, $right, $pivotIndex){ $pivotValue = $listArr[$pivotIndex]; swap($listArr[$pivotIndex],$listArr[$right]); $storeIndex = $left; echo "L: $left ,pval: $pivotValue ,right: $right <br />"; for ($i =$left;$i< $right;$i++){ if ($listArr[$i] < $pivotValue){ swap($listArr[$storeIndex] ,$listArr[$i]); //$listArr[$storeIndex] ^= $listArr[$i]^= $listArr[$storeIndex]^= $listArr[$i]; //This has a bug $storeIndex++; } } swap($listArr[$right],$listArr[$storeIndex]); // Move pivot to its final place var_dump($listArr); return $storeIndex; } //select k smallest value function select($listArr, $left, $right, $k){ $pivotIndex = floor(($right+$left)/2); $pivotNewIndex = partition($listArr, $left, $right, $pivotIndex); $pivotDist = $pivotNewIndex - $left + 1; //var_dump($listArr); //echo $pivotDist." ".$left.' '.$right." $k<br />"; if ($pivotDist == $k) return $listArr[$pivotNewIndex]; else if ($pivotDist>$k ) return select($listArr, $left, $pivotNewIndex - 1, $k); else return select($listArr, $pivotNewIndex + 1, $right, $k - $pivotDist); } $Arr=array(1,29,3,45,5,6,7,8,9,0,22,36,88,66,44,55,33,22,14,23,31,16,94,25,10,11,12,13,2,4); echo 'final result:'.select($Arr, 0, sizeof($Arr)-1,10);
N choose K in order
<?php $n=6; $k=3; $list=array(); for($i=0;$i<$n;$i++){ $list[$i]=$i; } function delete($list,$i){ $res=array(); for($j=0;$j<sizeof($list);$j++){ if($j>$i) array_push($res,$list[$j]); } //var_dump($res); return $res; } // 3/5 -> 2/4 -> 1/3 || 3/5 -> select 2nd 2/3-> 1/3 function select($list,$k,$head="") { $n=sizeof($list); //end condition if($n<$k) return; if($n==$k && $k>1){ echo $head.implode(' ',$list).'<br />'; } else if($k==1){ for($i=0;$i<$n;$i++) echo $head.$list[$i].'<br />'; } else if($n>$k){ for($i=0;$i<=$n-$k;$i++){//select the smallest element 4-3=1 //delete $i element and before in array $list1=delete($list,$i);//delete all elements before $i $head1=$head.$list[$i].' ';//construct head select($list1,$k-1,$head1);//select $k-1 from ($n-$i) } } } echo select($list,$k);
Max sub array
Kadane's algorithm
Kadane's algorithm consists of a scan through the array values, computing at each position the maximum subarray ending at that position. This subarray is either empty (in which case its sum is zero) or consists of one more element than the maximum subarray ending at the previous position. Thus, the problem can be solved with the following code, expressed here in Python:
def max_subarray(A): max_so_far = max_ending_here = 0 for x in A: max_ending_here = max(0, max_ending_here + x) max_so_far = max(max_so_far, max_ending_here) return max_so_far
Sub sum NP problem
Find a sub array which sum equals to 4.
input : {1,2,4,5,6,1,2,4,3,5,7,2,1}
output : {1,1,2}, {2,2}, {3,1}, {1,2,1}
<?php //Sub sum problem NP $arr=array(2,1,3,1,6,5,6,2,8,4); function subsum($arr,$sum,$head=""){ //find a possible num. if($sum<=0||sizeof($arr)==0) return; foreach($arr as $i=>$a){ if ($a>$sum){ unset($arr[$i]); continue; } else if($a==$sum){//this is the only possbility echo "Result: ".$head.$a." from sum $sum & head $head & a=$a<br />"; unset($arr[$i]); return; } else{ //keep $a $head1=$head.$a." "; unset($arr[$i]); // Unset only affects the current level of array. Does not affect the original outside this function. //echo "case3: keep a=$a and send to sum".($sum-$a)." with head=$head and arr_len=".sizeof($arr)."<br />"; subsum($arr,$sum-$a,$head1); //skip $a then continue } } return; } subsum($arr,5); /* output Result: 2 1 2 from sum 2 & head 2 1 & a=2 Result: 2 3 from sum 3 & head 2 & a=3 Result: 1 3 1 from sum 1 & head 1 3 & a=1 Result: 1 4 from sum 4 & head 1 & a=4 Result: 3 2 from sum 2 & head 3 & a=2 Result: 1 4 from sum 4 & head 1 & a=4 Result: 5 from sum 5 & head & a=5 */
Swap 2 numbers in one line
//swap $a=10; $b=20; $a=$a+$b-($b=$a); echo $a." and ". $b."\n"; // output 20 and 10 $a^=$b^=$a^=$b; //This has a bug. You cannot do things like this $a^=$a^=$a^=$a echo $a." and ". $b; //output 10 and 20
Calculate degree between hour and minute hand
<?php $time=array(); $time['h']=12; $time['m']=15; $time['s']=30; function clockDegree($time){ $mPercent=((float)$time['m']+((float)$time['s']/60.0))/60.0; $mDegree=$mPercent*360; $hdegree=((int)$time['h']%12+(float)$mPercent)/12.0*360; echo abs($mDegree-$hdegree); } clockDegree($time);
Permutation
A link to js fiddle http://jsfiddle.net/2v5fJ/
//this method is not very good, it requires some extra space var input_arr = [1,2,3,4,5,6]; var exclude_index = []; var prefix_arr = []; function perm(arr, exclude_index, prefix_arr) { if(prefix_arr.length == arr.length) { ///got an end console.log(prefix_arr); } else { for(var i = 0; i < arr.length; i++) { //select i from arr and place it in the new array if(exclude_index.indexOf(i) != -1) { continue; } var new_exclude_index = exclude_index.slice(0); new_exclude_index.push(i); var new_prefix_arr = prefix_arr.slice(0); new_prefix_arr.push(arr[i]); perm(arr, new_exclude_index, new_prefix_arr) ; } } } perm(input_arr, exclude_index, prefix_arr);