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My approach was different. I take a window from left to right and narrow the window from the left by one if the current sum is too large. (though I use subtraction rather than addition to compute this) This should be O(n) because, once we get to the end of the list we are done moving right and simply waiting for the tail pointer to get to the end (provided we are bigger than the sum when the index gets to the end)
- johnlabarge September 28, 2012