Algorithm

每日一题-A+B-

[每日一题] (A+B-)
给出两个字符串,s1和s2。s2中的+表示前面的字符重复2次，-代表前面的字符重复4次。问题是求s1中有多少个连续或不连续的s2。
例子: s1 = “aaabbbb” , s2 = “a+b-“
输出: 3

解题思路


使用动态规划。具体思路说起来有点绕…总之算是一个比较麻烦的动态规划题。

用n数组存储到当前位置为止有多少个n子串，用n_1数组存储到当前位置为止有多少个n-1子串
状态转移函数是n[i] = n[i-1] + n_1[i-1] + n_1[i-2] + … + n_1[2]或n_1[4]，大概就是这个意思

代码没有经过很多测试，可能会有bug

Python:

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#!/usr/bin/env python
# -*- coding:utf-8 -*-
#s1 = "aabbbbaacc"
#s2 = "a+b-c+"
#s1 = "aaabababbaabbccc"
#s2 = "a+b-c+"
#s1 = "aababaaabaaab"
#s1 = "aabaabaabcc"
#s2 = "a+b+c+"
#s1 = "waeginsapnaabangpisebbasepgnccccapisdnfngaabndlrjngeuiogbbegbuoecccc"
#s2 = "a+b+c-"
s1 = "wweqwqaabddakswbbcc"
s2 = "a+b+"
class Solution(object):
    def aPlusbPlus(self, s1, s2):
        pre = {} # 记录当前字符的前一字符
        index = {} # 记录当前字符位置
        count = {} # 记录当前字符找了几个
        bound = {} # 记录当前字符要找几个
        p = None
        for i in xrange(len(s2)/2):
            pre[s2[i*2]] = p
            index[s2[i*2]] = 0
            count[s2[i*2]] = 0
            bound[s2[i*2]] = 2 if s2[i*2+1] == '+' else 4
            p = s2[i*2]
        n_1 = [[0, 0] for i in xrange(len(s1))]
        n = [[0, 0] for i in xrange(len(s1))]
        for i in xrange(len(s1)):
            c = s1[i]
            if c in index.keys():
                if c == s2[0]: # 当前字符为s2首字符
                    count[c] += 1
                    sum = 1
                    for j in xrange(bound[c]):
                        sum *= count[c] - j
                    n[i] = [sum / bound[c], c]
                    n_1[i] = [0, c]
                    index[c] = i
                else:
                    count[c] += 1
                    n_1[i] = [n[index[pre[c]]][0], c]
                    if count[c] < bound[c]:
                        n[i] = [0, c]
                    else :
                        n[i] = [n[index[c]][0], c]
                        count1, j = count[c] - 1, i-1
                        while count1 >= bound[c]-1 and j >= 0:
                            if n_1[j][1] == c:
                                n[i][0] += n_1[j][0]
                                count1 -= 1
                            j -= 1
                    index[c] = i
        return n[index[s2[len(s2)-2]]][0]