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社区首页 >专栏 >leetcode: 38. Count and Say

leetcode: 38. Count and Say

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JNingWei
发布2018-09-27 17:08:23
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发布2018-09-27 17:08:23
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文章被收录于专栏:JNing的专栏JNing的专栏

Problem

# The count-and-say sequence is the sequence of integers 
# with the first five terms as following:
#
# 1.     1
# 2.     11
# 3.     21
# 4.     1211
# 5.     111221
#
# 1 is read off as "one 1" or 11.
# 11 is read off as "two 1s" or 21.
# 21 is read off as "one 2, then one 1" or 1211.
#
# Given an integer n, generate the nth term of the count-and-say sequence.
#
# Note: Each term of the sequence of integers will be represented as a string.
#
# Example 1:
#
# Input: 1
# Output: "1"
#
# Example 2:
#
# Input: 4
# Output: "1211"

AC

class Solution():
    def countAndSay(self, n):
        if not n:
            return ''
        res = "1"
        for _ in range(n-1):
            tmp, count = '', 1
            for i in range(1, len(res)):
                if res[i] == res[i-1]:
                    count += 1
                else:
                    tmp += (str(count) + str(res[i-1]))
                    count = 1
            tmp += (str(count) + str(res[-1]))
            res = tmp
        return res


if __name__ == "__main__":
    assert Solution().countAndSay(1) == '1'
    assert Solution().countAndSay(2) == '11'
    assert Solution().countAndSay(3) == '21'
    assert Solution().countAndSay(4) == '1211'
    assert Solution().countAndSay(5) == '111221'
    assert Solution().countAndSay(6) == '312211'
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