Given two strings A and B of lowercase letters, return true if and only if we can swap two letters in A so that the result equals B.
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Input: A = "ab", B = "ba"
Output: true
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-
Example 2:
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Input: A = "ab", B = "ab"
Output: false
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Example 3:
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Input: A = "aa", B = "aa"
Output: true
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Example 4:
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Input: A = "aaaaaaabc", B = "aaaaaaacb"
Output: true
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Input: A = "", B = "aa"
Output: false
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Note:
- 0 <= A.length <= 20000
- 0 <= B.length <= 20000
- A and B consist only of lowercase letters.
交换A字符串的两个字符，使得 A = B.
实现
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public boolean (String A, String B) {
if (A.length() != B.length() ) {
return false;
}
if (A.length() < 2 || B.length() < 2){
return false;
}
int len = A.length();
char[] chA = A.toCharArray();
char[] chB = B.toCharArray();
int modCount = 0;
char[] ele = new char[4];
Set<Character> keySet = new HashSet<>();
for (int i = 0; i < len; i++) {
char a = chA[i];
char b = chB[i];
keySet.add(a);
if (modCount > 2){
return false;
}
if (a != b) {
modCount += 1;
if (ele[0] == 'u0000'){
ele[0] = a;
ele[1] = b;
} else {
ele[2] = a;
ele[3] = b;
}
}
}
if (modCount == 0){
return keySet.size() < chA.length;
}
return ele[0] == ele[3] && ele[1] == ele[2];
}
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讨论区其他答案
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public boolean (String A, String B) {
if (A.length() != B.length()){
return false;
}
if (A.equals(B)) {
Set<Character> s = new HashSet<Character>();
for (char c : A.toCharArray()) {
s.add(c);
}
return s.size() < A.length();
}
List<Integer> dif = new ArrayList<>();
for (int i = 0; i < A.length(); ++i) if (A.charAt(i) != B.charAt(i)) dif.add(i);
return dif.size() == 2 && A.charAt(dif.get(0)) == B.charAt(dif.get(1)) && A.charAt(dif.get(1)) == B.charAt(dif.get(0));
}
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参考资料
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