Lazily compute longestCommonX

[make-github-pseudonymous-again]

Currently there are still some extensions computation that could be avoided. For instance, here

myers-1986/src/twoWayScan.js

Lines 47 to 53 in da6d7e5

	for (const k of forwardStep(centerF, D, V, eq, li, lj, ri, rj, Delta0)) {
	const x = V[centerF + k];
	const y = x - (k + Delta0);
	V[centerF + k] = longestCommonPrefix(eq, x, lj, y, rj);
	if (k - Delta < -(D - parityDelta)) continue;
	if (k - Delta > D - parityDelta) continue;
	if (V[centerF + k] < V[centerB + k - Delta]) continue; // TODO this scans the snake twice

The value of V[centerF + k] is not needed until the end of the forwardStep loop if the k - Delta bound checks fail.

[make-github-pseudonymous-again]

One quick solution would be to mirror backwardExtend as forwardExtend, move the computation of V[centerF + k] until after the bound checks, and call forwardExtend once the forwardStep loop is done.