Tor Hedin Brønner 4d9c75ca71 Sort the stack by distance and then mru
O(`max`*`space.length`) ~ O(`space.length`) for finitely wide screens. Doesn't
seem like there's a large performance penalty to decide ties between
equidistant windows using the mru, instead of deciding ties by sorting on
left/right.
2018-02-17 21:16:27 +01:00
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GPL-3.0
2 MiB
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