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Sign up【每日一题】- 2019-11-11 859.亲密字符串 #229
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python3实现 class Solution:
def buddyStrings(self, A: str, B: str) -> bool:
s1 = ''
s2 = ''
if len(A) != len(B):
return False
elif A == B:
if len(set(A)) < len(A):
return True
else:
return False
else:
for i in range(len(A)):
if A[i] != B[i]:
s1 += A[i]
s2 += B[i]
if s1[0] == s2[1] and s1[1] == s2[0] and len(s1)==2:
return True
else:
return False |
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class Solution {
public boolean buddyStrings(String A, String B) {
int lenA = A.length();
int lenB = B.length();
if(lenA < 2 || lenA != lenB) {
return false;
}
int diffCount = 0;
char char1 = 0;
char char2 = 0;
HashSet<Character> charSet = new HashSet<>();;
for(int i = 0; i < lenA; i++) {
charSet.add(A.charAt(i));
if(A.charAt(i) != B.charAt(i)) {
if(diffCount == 0) {
char1 = A.charAt(i);
char2 = B.charAt(i);
}
else if(diffCount == 1) {
if(char1 != B.charAt(i) || char2 != A.charAt(i)) {
return false;
}
}
diffCount++;
}
}
return diffCount == 2 || charSet.size() < lenA;
}
} |
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C 实现: void Swap(char* pCh1, char* pCh2) BOOL buddyStrings(CHAR pStrA[], CHAR pStrB[]) |
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bool isBuddyString(string str1, string str2) { |
给定两个由小写字母构成的字符串 A 和 B ,只要我们可以通过交换 A 中的两个字母得到与 B 相等的结果,就返回 true ;否则返回 false 。
示例 1:
输入: A = "ab", B = "ba"
输出: true
示例 2:
输入: A = "ab", B = "ab"
输出: false
示例 3:
输入: A = "aa", B = "aa"
输出: true
示例 4:
输入: A = "aaaaaaabc", B = "aaaaaaacb"
输出: true
示例 5:
输入: A = "", B = "aa"
输出: false
提示:
0 <= A.length <= 20000
0 <= B.length <= 20000
A 和 B 仅由小写字母构成。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/buddy-strings
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