Leetcode: 【每日一题】- 2019-11-11 859.亲密字符串

[azl397985856]

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
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

[wangyuxiang0829]

class Solution {
public:
    bool buddyStrings(string A, string B) {
        int m = A.size(), n = B.size(), c = 0;
        if (m <= 1 || m != n) return false;
        vector<int> v;
        int map[26] = {0};
        bool flag = false;
        for (int i = 0; i < n; i++) {
            if (int m = ++map[A[i] - 'a']; m >= 2)
                flag = true;
            if (A[i] != B[i]) {
                c++;
                if (c > 2) return false;
                v.push_back(i);
            }
        }
        if (c == 1) return false;
        else if (c == 0) return flag;
        else return A[v[0]] == B[v[1]] && A[v[1]] == B[v[0]];
    }
};

[Stephen-Pierre]

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

[wptercel]

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;
    }
}

[XRuHa]

C 实现：

void Swap(char* pCh1, char* pCh2)
{
char temp = 0;
temp = *pCh1;
*pCh1 = *pCh2;
*pCh2 = temp;
}

BOOL buddyStrings(CHAR pStrA[], CHAR pStrB[])
{
size_t length = strlen(pStrA);
BOOL blFlag = FALSE;
for (size_t i = 1; i < length; i++)
{
Swap(&pStrA[i-1], &pStrA[i]);
if (0 == strcmp(pStrA, pStrB))
{
blFlag = TRUE;
break;
}
}
return blFlag ? TRUE : FALSE;
}

[Casper-Mars]

bool isBuddyString(string str1, string str2) {
/判断是否为空和长度是否相等/
if (str1.empty() || str2.empty() || str1.size() != str2.size()) {
return false;
}
string::iterator str1b = str1.begin();
string::iterator str2b = str2.begin();
int swapTime = 0;
/遍历字符串/
while (str1b != str1.end()) {
/如果对应位置不相同/
if (str1b != *str2b) {
/两个字符串当前位置和下一位置的交叉比对，相同则记录交换次数，不同则直接返回false/
if ((str1b + 1) == str2b && *(str2b + 1) == *str1b) {
/交换次数大于1次返回false*/
if (swapTime >= 1) {
return false;
} else {
swapTime++;
str1b++;
str2b++;
}
} else {
return false;
}
}
str1b++;
str2b++;
}
return true;
}

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