class Solution {
public:
bool buddyStrings(string s, string goal) {
}
};
859. 亲密字符串
给你两个字符串 s 和 goal ，只要我们可以通过交换 s 中的两个字母得到与 goal 相等的结果，就返回 true ；否则返回 false 。
交换字母的定义是：取两个下标 i 和 j （下标从 0 开始）且满足 i != j ，接着交换 s[i] 和 s[j] 处的字符。
"abcd" 中交换下标 0 和下标 2 的元素可以生成 "cbad" 。
示例 1：
输入：s = "ab", goal = "ba" 输出：true 解释：你可以交换 s[0] = 'a' 和 s[1] = 'b' 生成 "ba"，此时 s 和 goal 相等。
示例 2：
输入：s = "ab", goal = "ab" 输出：false 解释：你只能交换 s[0] = 'a' 和 s[1] = 'b' 生成 "ba"，此时 s 和 goal 不相等。
示例 3：
输入：s = "aa", goal = "aa" 输出：true 解释：你可以交换 s[0] = 'a' 和 s[1] = 'a' 生成 "aa"，此时 s 和 goal 相等。
提示：
1 <= s.length, goal.length <= 2 * 104s 和 goal 由小写英文字母组成原站题解
javascript 解法, 执行用时: 64 ms, 内存消耗: 41.9 MB, 提交时间: 2023-09-27 14:50:28
/**
* @param {string} s
* @param {string} goal
* @return {boolean}
*/
var buddyStrings = function(s, goal) {
if (s.length != goal.length) {
return false;
}
if (s === goal) {
const count = new Array(26).fill(0);
for (let i = 0; i < s.length; i++) {
count[s[i].charCodeAt() - 'a'.charCodeAt()]++;
if (count[s[i].charCodeAt() - 'a'.charCodeAt()] > 1) {
return true;
}
}
return false;
} else {
let first = -1, second = -1;
for (let i = 0; i < s.length; i++) {
if (s[i] !== goal[i]) {
if (first === -1)
first = i;
else if (second === -1)
second = i;
else
return false;
}
}
return (second !== -1 && s[first] === goal[second] && s[second] === goal[first]);
}
};
python3 解法, 执行用时: 32 ms, 内存消耗: 16.1 MB, 提交时间: 2023-09-27 14:50:12
class Solution:
def buddyStrings(self, s: str, goal: str) -> bool:
if len(s) != len(goal):
return False
if s == goal:
if len(set(s)) < len(goal):
return True
else:
return False
diff = [(a, b) for a, b in zip(s, goal) if a != b]
return len(diff) == 2 and diff[0][0] == diff[1][1] and diff[0][1] == diff[1][0]
java 解法, 执行用时: 1 ms, 内存消耗: 40.5 MB, 提交时间: 2023-09-27 14:49:57
class Solution {
public boolean buddyStrings(String s, String goal) {
if (s.length() != goal.length()) {
return false;
}
if (s.equals(goal)) {
int[] count = new int[26];
for (int i = 0; i < s.length(); i++) {
count[s.charAt(i) - 'a']++;
if (count[s.charAt(i) - 'a'] > 1) {
return true;
}
}
return false;
} else {
int first = -1, second = -1;
for (int i = 0; i < goal.length(); i++) {
if (s.charAt(i) != goal.charAt(i)) {
if (first == -1)
first = i;
else if (second == -1)
second = i;
else
return false;
}
}
return (second != -1 && s.charAt(first) == goal.charAt(second) &&
s.charAt(second) == goal.charAt(first));
}
}
}
cpp 解法, 执行用时: 4 ms, 内存消耗: 7.2 MB, 提交时间: 2023-09-27 14:49:43
class Solution {
public:
bool buddyStrings(string s, string goal) {
if (s.size() != goal.size()) {
return false;
}
if (s == goal) {
vector<int> count(26);
for (int i = 0; i < s.size(); i++) {
count[s[i] - 'a']++;
if (count[s[i] - 'a'] > 1) {
return true;
}
}
return false;
} else {
int first = -1, second = -1;
for (int i = 0; i < s.size(); i++) {
if (s[i] != goal[i]) {
if (first == -1)
first = i;
else if (second == -1)
second = i;
else
return false;
}
}
return (second != -1 && s[first] == goal[second] && s[second] == goal[first]);
}
}
};
golang 解法, 执行用时: 0 ms, 内存消耗: 2.2 MB, 提交时间: 2023-09-27 14:49:19
func buddyStrings(s, goal string) bool {
if len(s) != len(goal) {
return false
}
if s == goal {
seen := [26]bool{}
for _, ch := range s {
if seen[ch-'a'] {
return true
}
seen[ch-'a'] = true
}
return false
}
first, second := -1, -1
for i := range s {
if s[i] != goal[i] {
if first == -1 {
first = i
} else if second == -1 {
second = i
} else {
return false
}
}
}
return second != -1 && s[first] == goal[second] && s[second] == goal[first]
}
golang 解法, 执行用时: 0 ms, 内存消耗: 2.4 MB, 提交时间: 2021-06-07 16:55:35
func buddyStrings(s string, goal string) bool {
if len(s) != len(goal) {
return false
}
if s == goal {
mp := map[byte]int{}
for i := 0; i < len(s); i++ {
if _, ok := mp[s[i]]; ok {
return true
} else {
mp[s[i]]++
}
}
return false
}
var pair [][]byte
for i := 0; i < len(s); i++ {
if s[i] != goal[i] {
pair = append(pair, []byte{s[i], goal[i]})
}
if len(pair) > 2 {
return false
}
}
if len(pair) != 2 {
return false
}
return pair[0][0] == pair[1][1] && pair[0][1] == pair[1][0]
}