91. Decode Ways

作者: oo上海 | 来源:发表于2016-07-19 07:24 被阅读15次

    91. Decode Ways

    题目:
    https://leetcode.com/problems/decode-ways/

    tag : DP

    难度 : Medium

    BASE CASE(len(s) = 1 和 len(s) = 2 ): 
    直接check
    
    非BASE CASE :
    先令 dp[i] = 0
    如果s[i]是可以map的话 -> dp[i] += dp[i-1] 原本的s[0..i]decode方式加上s[i]
    如果s[i-1,i]可以map的话 -> dp[i] += dp[i-2] 原本的s[0...i-1]decode方式加上s[i-1,i]
    

    Python代码(可美化)

    class Solution(object):
        def numDecodings(self, s):
            """
            :type s: str
            :rtype: int
            """
            keys = ['1', '2', '3', '4', '5', '6', '7', '8', '9' ,'10', '11', '12', '13', '14', '15', '16', '17', '18', '19', '20', '21', '22', '23', '24', '25', '26']
            values = ['A', 'B','C', 'D', 'E', 'F', 'G','H', 'I', 'J', 'K', 'L', 'M' , 'N', 'O', 'P','Q', 'S', 'R', 'T', 'U','V', 'W', 'X','Y','Z']
            numbersToLetters = dict(zip(keys, values))
        
            ways = {}
            n = len(s)
            for i in range(n):
                ways[i] = 0
            if n == 0:
                return 0
            elif n == 1 :
                ways[0] = int(s in numbersToLetters)
            elif n == 2:
                if (s[0] in numbersToLetters) and (s[1] in numbersToLetters):
                    ways[1] += 1
                if (s in numbersToLetters):
                    ways[1] += 1
            else:
                #s[0]
                ways[0] = int(s[0] in numbersToLetters)
                #s[01]
                if (s[0] in numbersToLetters) and (s[1] in numbersToLetters):
                    ways[1] += 1
                if (s[:2] in numbersToLetters):
                    ways[1] += 1        
                for i in range(2,n):
                    if s[i] in numbersToLetters:
                        ways[i] += ways[i-1]
                    if (s[i-1:i+1] in numbersToLetters):
                        ways[i] += ways[i-2]
                
            #print(ways[n-1])
            return ways[n-1]
                
    

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