how can I simplify this code?

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Cesar Cardenas
Cesar Cardenas 2025-9-13
回答： Chuguang Pan 2025-9-14
Ran in:
is there any way to simplify this code? any help would be appreciated thanks.
function [idx, pairs_checked, scan_indices] = two_ended_search_1(x, target)
% TWO_ENDED_SEARCH Bidirectional scan from array ends toward center (arrays-only).
n = length(x);
% --- Edge case: empty array ---
if n == 0
idx = -1;
pairs_checked = zeros(0,2);
scan_indices = zeros(1,0);
return;
end
% --- Preallocate with conservative sizes ---
maxIters = ceil(n/2);
pairs_checked = zeros(maxIters, 2);
scan_indices = zeros(1, 2*maxIters); % at most two comparisons per iter
L = 1; R = n;
idx = -1;
pairCount = 0;
scanCount = 0;
while L <= R
% Record pair at start of iteration
pairCount = pairCount + 1;
pairs_checked(pairCount, :) = [L, R];
% Compare left
scanCount = scanCount + 1;
scan_indices(scanCount) = L;
if x(L) == target
idx = L;
break;
end
% If middle element already checked, stop (avoid duplicate compare)
if L == R
break;
end
% Compare right
scanCount = scanCount + 1;
scan_indices(scanCount) = R;
if x(R) == target
idx = R;
break;
end
% Move pointers inward
L = L + 1;
R = R - 1;
end
% Trim prealloc arrays to actual sizes
pairs_checked = pairs_checked(1:pairCount, :);
scan_indices = scan_indices(1:scanCount);
end
x = [5, 2, 3, 1, 9, 7, 9, 4, 8];
target = 4;
[idx, pairs_checked, scan_indices] = two_ended_search_1(x, target)
idx = 8
pairs_checked = 2×2
1 9 2 8
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scan_indices = 1×4
1 9 2 8
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采纳的回答

Chuguang Pan
Chuguang Pan 2025-9-14
Ran in:
function [idx,pairs,scan] = Two_Ended_Search_2(x,target)
idx = find(x==target);
n = length(x);
LR = min(idx-1,n-idx);
pairsL = 1:(LR+1);
pairsR = n:-1:n-LR;
pairs = [pairsL.' pairsR.'];
scan = reshape(pairs.',1,[]);
end
X = [5, 2, 3, 1, 9, 7, 9, 4, 8];
target = 4;
[Idx,Paris,Scan] = Two_Ended_Search_2(X,target)
Idx = 8
Paris = 2×2
1 9 2 8
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Scan = 1×4
1 9 2 8
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