Slightly optimize Graphs.oppositeNode(). 50% of the time (assuming node is randomly chosen between the repeated node and adjacent node) we'll only have to perform one equals() check instead of two. ------------- Created by MOE: https://github.com/google/moe MOE_MIGRATED_REVID=114380996
Slightly optimize Graphs.oppositeNode(). 50% of the time (assuming node is randomly chosen between the repeated node and adjacent node) we'll only have to perform one equals() check instead of two.
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Created by MOE: https://github.com/google/moe
MOE_MIGRATED_REVID=114380996

google/guava,berndhopp/guava,DavesMan/guava,google/guava,xasx/guava,rgoldberg/guava,jakubmalek/guava,xasx/guava,berndhopp/guava,google/guava,typetools/guava,tli2/guava,mosoft521/guava,mosoft521/guava,mosoft521/guava,rgoldberg/guava,tli2/guava,Xaerxess/guava,EdwardLee03/guava,EdwardLee03/guava,DavesMan/guava,jakubmalek/guava,typetools/guava,Xaerxess/guava