给定两个字符串 s
和 t
,它们只包含小写字母。
字符串 t
由字符串 s
随机重排,然后在随机位置添加一个字母。
请找出在 t
中被添加的字母。
示例 1:
输入:s = "abcd", t = "abcde" 输出:"e" 解释:'e' 是那个被添加的字母。
示例 2:
输入:s = "", t = "y" 输出:"y"
提示:
0 <= s.length <= 1000
t.length == s.length + 1
s
和t
只包含小写字母
方法一:计数
使用数组(cnt
)统计 s
与 t
当中字符出现的次数:s[i]
进行 cnt[s[i] - 'a']++
,t[i]
进行 cnt[t[i] - 'a']--
。
完成统计后,找到符合 cnt[i] == -1
的 i
,返回即可(return 'a' + i
)。
时间复杂度
方法二:求和
由于 s
与 t
只存在一个不同元素,可以统计两者所有字符 ASCII 码之和,再进行相减(sum(t) - sum(s)
),即可得到 t
中那一个额外字符的 ASCII 码。
时间复杂度
class Solution:
def findTheDifference(self, s: str, t: str) -> str:
cnt = Counter(s)
for c in t:
cnt[c] -= 1
if cnt[c] < 0:
return c
class Solution:
def findTheDifference(self, s: str, t: str) -> str:
a = sum(ord(c) for c in s)
b = sum(ord(c) for c in t)
return chr(b - a)
class Solution {
public char findTheDifference(String s, String t) {
int[] cnt = new int[26];
for (int i = 0; i < s.length(); ++i) {
++cnt[s.charAt(i) - 'a'];
}
for (int i = 0; ; ++i) {
if (--cnt[t.charAt(i) - 'a'] < 0) {
return t.charAt(i);
}
}
}
}
class Solution {
public char findTheDifference(String s, String t) {
int ss = 0;
for (int i = 0; i < t.length(); ++i) {
ss += t.charAt(i);
}
for (int i = 0; i < s.length(); ++i) {
ss -= s.charAt(i);
}
return (char) ss;
}
}
class Solution {
public:
char findTheDifference(string s, string t) {
int cnt[26] = {0};
for (char& c : s) ++cnt[c - 'a'];
for (char& c : t) if (--cnt[c - 'a'] < 0) return c;
return ' ';
}
};
class Solution {
public:
char findTheDifference(string s, string t) {
int a = 0, b = 0;
for (char& c : s) a += c;
for (char& c : t) b += c;
return b - a;
}
};
function findTheDifference(s: string, t: string): string {
const n = s.length;
const count = new Array(26).fill(0);
for (let i = 0; i < n; i++) {
count[s.charCodeAt(i) - 'a'.charCodeAt(0)]++;
count[t.charCodeAt(i) - 'a'.charCodeAt(0)]--;
}
count[t.charCodeAt(n) - 'a'.charCodeAt(0)]--;
return String.fromCharCode(
'a'.charCodeAt(0) + count.findIndex(v => v !== 0),
);
}
function findTheDifference(s: string, t: string): string {
return String.fromCharCode(
[...t].reduce((r, v) => r + v.charCodeAt(0), 0) -
[...s].reduce((r, v) => r + v.charCodeAt(0), 0),
);
}
impl Solution {
pub fn find_the_difference(s: String, t: String) -> char {
let s = s.as_bytes();
let t = t.as_bytes();
let n = s.len();
let mut count = [0; 26];
for i in 0..n {
count[(s[i] - b'a') as usize] += 1;
count[(t[i] - b'a') as usize] -= 1;
}
count[(t[n] - b'a') as usize] -= 1;
char::from(b'a' + count.iter().position(|&v| v != 0).unwrap() as u8)
}
}
impl Solution {
pub fn find_the_difference(s: String, t: String) -> char {
let mut ans = 0;
for c in s.as_bytes() {
ans ^= c;
}
for c in t.as_bytes() {
ans ^= c;
}
char::from(ans)
}
}
char findTheDifference(char *s, char *t) {
int n = strlen(s);
int count[26] = {0};
for (int i = 0; i < n; i++) {
count[s[i] - 'a']++;
count[t[i] - 'a']--;
}
count[t[n] - 'a']--;
int i;
for (i = 0; i < 26; i++) {
if (count[i]) {
break;
}
}
return 'a' + i;
}
char findTheDifference(char *s, char *t) {
int n = strlen(s);
char ans = 0;
for (int i = 0; i < n; i++) {
ans ^= s[i];
ans ^= t[i];
}
ans ^= t[n];
return ans;
}
func findTheDifference(s, t string) byte {
cnt := [26]int{}
for _, ch := range s {
cnt[ch-'a']++
}
for i := 0; ; i++ {
ch := t[i]
cnt[ch-'a']--
if cnt[ch-'a'] < 0 {
return ch
}
}
}
func findTheDifference(s string, t string) byte {
ss := 0
for _, c := range s {
ss -= int(c)
}
for _, c := range t {
ss += int(c)
}
return byte(ss)
}