这是一个简单的单遍O（n）解决方案：

s = [1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11,42]
maxrun = -1
rl = {}
for x in s:
    run = rl[x] = rl.get(x-1, 0) + 1
    print x-run+1, 'to', x
    if run > maxrun:
        maxend, maxrun = x, run
print range(maxend-maxrun+1, maxend+1)

如果您考虑范围而不是端点和游程长度的单个变量，那么逻辑可能会更加不言而喻：

rl = {}
best_range = xrange(0)
for x in s:
    run = rl[x] = rl.get(x-1, 0) + 1
    r = xrange(x-run+1, x+1)
    if len(r) > len(best_range):
        best_range = r
print list(best_range)