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  • 【leetcode】522. Longest Uncommon Subsequence II

    题目如下:

    解题思路:因为given list长度最多是50,我的解法就比较随意了,直接用一个嵌套的循环,判断数组中每个元素是否是其他的subsequence,最后找出不属于任何元素subsequence的最长元素即可。

    代码如下:

    class Solution(object):
        def isSubsequence(self, a, b):
            """
            :type a: str
            :type b: str
            :rtype: int
            """
            s = b
            for i in a:
                inx = s.find(i)
                if inx == -1:
                    return False
                s = s[inx+1:]
            return True
    
        def findLUSlength(self, strs):
            """
            :type strs: List[str]
            :rtype: int
            """
            def cmpf(v1,v2):
                return len(v2) - len(v1)
            strs.sort(cmp = cmpf)
            if len(strs) == 0 or len(strs[0]) == 0:
                return -1
            res = -1
    
            visit = [0 for x in strs]
            for i in range(len(strs)):
                if visit[i] == 1:
                    continue
                tmp = False
                for j in range(len((strs))):
                    if i == j:
                        continue
                    tmp = self.isSubsequence(strs[i],strs[j])
                    if tmp == True:
                        visit[i] = 1
    
            for i in xrange(len(visit)):
                if visit[i] == 0:
                    return len(strs[i])
            return -1
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  • 原文地址:https://www.cnblogs.com/seyjs/p/9143530.html
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