LeetCode | 023.02.27 | 1638. 統計只差一個字符的子串數目

"""
題目分析:
    這是中等難度題，讀題目都沒讀懂；看了題解中的說明才理解題目的意思；實際上就是求字符串s和字符串t中有多少子字符串滿足：只有一位不同
理論:暴力求解
    1.分別循環s和t
    2.每次循環前，初始化diff和指針cur
    3.循環當前指針cur，檢查當個s和t子字符串
    4.如果當前s和t的不同，diff+1
    5.1如果diff==1，表示只有一個不同，加入結果ret+1
    5.2如果diff>1，表示不滿足條件，直接break
    6.返回結果ret
注意/難點：
    s的子字符串和t的子字符串長度相同
"""
class Solution:
    def countSubstrings(self, s: str, t: str) -> int:
        ret=0                                                #初始化變量
        for i in range(len(s)):                              #循環s
            for j in range(len(t)):                          #循環t
                diff=0                                       #字符差異
                cur=0                                        #當前指針
                s1,s2="",""                    #debug only
                while cur+i<len(s) and cur+j<len(t):         #移動當前指針
                    s1+=s[cur+i]               #debug only
                    s2+=t[cur+j]               #debug only
                    if s[cur+i]!=t[cur+j]:                   #檢查是否相同k
                        diff+=1                              #不同數
                    if diff==1:                              #有一個不同
                        ret+=1                               #加入結果
                        print("s:"+s1,"t:"+s2)
                    elif diff>1:                              #>一個不同
                        break                                #退出循環
                    cur+=1                                   #當前指針後移
        return ret                                           #返回結果
        

s = "aba"
t = "baba"
ans=Solution().countSubstrings(s,t)
print(ans)

輸出的結果如下:

以下是用電子表格模擬變量的部分過程:

			s=[cur+i];t=[cur+j]		diff	ret
循環s			i=0	a
	循環t		j=0	b
		循環當前指針cur	cur=0		0	0
		判斷	s[0+0]=s[0+0]	a<>b
		判斷	diff		1	1
	t+1		j=1	a
		循環當前指針cur	cur=0		0	0
		判斷	s[0+0]=s[0+1]	a=a
		判斷	diff		0	0
		循環當前指針cur	cur=1		0	0
		判斷	s[1+0]=s[1+1]	b=b
		判斷	diff		0	0
		循環當前指針cur	cur=2		0	0
		判斷	s[2+0]=s[2+1]	a=a
		判斷	diff		0	0
	t+1		j=2	a
		循環當前指針cur	cur=0		0	0
		判斷	s[0+0]=s[0+2]	a<>b
		判斷	diff		1	1
		循環當前指針cur	cur=1		0	0
		判斷	s[1+0]=s[1+1]	b=b
		判斷	diff		0	0
		循環當前指針cur	cur=2		0	0
		判斷	s[2+0]=s[2+1]	a=a
		判斷	diff		0	0
	t+1		j=3	a
		循環當前指針cur	cur=0		0	0
		判斷	s[0+0]=s[0+3]	a=a
		判斷	diff		0	0
		循環當前指針cur	cur=1		0	0
		判斷	s[1+0]=s[1+3]	退出循環
s+1			i=1	b
	循環t		j=0	b
		循環當前指針cur	cur=0		0	0
		判斷	s[0+1]=s[0+0]	b=b
		判斷	diff		0	0
	t+1		j=1	a
		循環當前指針cur	cur=0		0	0
		判斷	s[0+1]=s[0+1]	b<>a
		判斷	diff		1	1
		循環當前指針cur	cur=1		0	0
		判斷	s[1+1]=s[1+1]	a<>b
		判斷	diff		1	2
			退出循環
	t+1		j=2	a
		循環當前指針cur	cur=0		0	0
		判斷	s[0+1]=s[0+2]	b=b
		判斷	diff		0	0
		循環當前指針cur	cur=1		0	0
		判斷	s[1+1]=s[1+2]	a=a
		判斷	diff		0	0
			退出循環
	t+1		j=3	a
		循環當前指針cur	cur=0		0	0
		判斷	s[0+1]=s[0+3]	b<>a
		判斷	diff		1	1
		循環當前指針cur	cur=1		0	0
		判斷	s[1+0]=s[1+3]	退出循環