def incSeq(seq): start = 0 for i in xrange(1, len(seq)): if seq[i] < seq[i-1]: yield start, i - start start = i maxIncSeq = reduce(lambda x,y: x if x[1]>y[1] else y, incSeq(seq))
得到最长递增子串长度及起始位置,时间复杂度O(n).