2531.使字符串中不同字符的数目相等
链接：2531.使字符串中不同字符的数目相等
难度：Medium
标签：哈希表、字符串、计数
简介：给你两个下标从 0 开始的字符串 word1 和 word2 。如果可以通过 恰好一次 移动，使 word1 和 word2 中不同字符的数目相等，则返回 true ；否则，返回 false 。
题解 1 - typescript
- 编辑时间：2023-01-08
- 执行用时：160ms
- 内存消耗：49MB
- 编程语言：typescript
- 解法介绍：存储后遍历。
const list = new Array(26).fill(0).map((_, i) => String.fromCodePoint(i + 97));
function isItPossible(word1: string, word2: string): boolean {
const map1 = new Map<string, number>();
for (const c of word1) map1.set(c, (map1.get(c) ?? 0) + 1);
const map2 = new Map<string, number>();
for (const c of word2) map2.set(c, (map2.get(c) ?? 0) + 1);
const len1 = map1.size;
const len2 = map2.size;
for (const c1 of list) {
for (const c2 of list) {
if (check(c1, c2)) return true;
}
}
return false;
function check(c1: string, c2: string): boolean {
const cnt1 = map1.get(c1) ?? 0;
const t1 = map1.get(c2) ?? 0;
const cnt2 = map2.get(c2) ?? 0;
const t2 = map2.get(c1) ?? 0;
if (cnt1 === 0 || cnt2 === 0) return false;
if (c1 === c2) return len1 === len2;
if (cnt1 === 1 && cnt2 === 1) {
if (t1 === 0 && t2 === 0) return len1 === len2;
if (t1 === 0 && t2 !== 0) return len1 + 1 === len2;
if (t1 !== 0 && t2 === 0) return len1 === len2 + 1;
if (t1 !== 0 && t2 !== 0) return len1 === len2;
}
if (cnt1 === 1 && cnt2 > 1) {
if (t1 === 0 && t2 === 0) return len1 === len2 + 1;
if (t1 === 0 && t2 !== 0) return len1 === len2;
if (t1 !== 0 && t2 === 0) return len1 - 1 === len2 + 1;
if (t1 !== 0 && t2 !== 0) return len1 - 1 === len2;
}
if (cnt1 > 1 && cnt2 === 1) {
if (t1 === 0 && t2 === 0) return len2 === len1 + 1;
if (t1 !== 0 && t2 === 0) return len2 === len1;
if (t1 === 0 && t2 !== 0) return len2 - 1 === len1 + 1;
if (t1 !== 0 && t2 !== 0) return len2 - 1 === len1;
}
if (cnt1 > 1 && cnt2 > 1) {
if (t1 === 0 && t2 === 0) return len1 === len2;
if (t1 !== 0 && t2 === 0) return len1 === len2 + 1;
if (t1 === 0 && t2 !== 0) return len1 + 1 === len2;
if (t1 !== 0 && t2 !== 0) return len1 === len2;
}
return false;
}
}
题解 2 - rust
- 编辑时间：2023-01-08
- 执行用时：4ms
- 内存消耗：2.4MB
- 编程语言：rust
- 解法介绍：同上, 优化。
impl Solution {
pub fn is_it_possible(word1: String, word2: String) -> bool {
let mut list = [0; 26];
for i in 0..26 {
list[i] = i;
}
let (mut m1, mut len1) = ([0; 26], 0);
let (mut m2, mut len2) = ([0; 26], 0);
for c in word1.chars() {
let idx = c as usize - 'a' as usize;
m1[idx] += 1;
if m1[idx] == 1 {
len1 += 1;
}
}
for c in word2.chars() {
let idx = c as usize - 'a' as usize;
m2[idx] += 1;
if m2[idx] == 1 {
len2 += 1;
}
}
for i1 in list.iter() {
for i2 in list.iter() {
let (i1, i2) = (*i1, *i2);
let (c1, t1) = (m1[i1], m1[i2]);
let (c2, t2) = (m2[i2], m2[i1]);
if c1 == 0 || c2 == 0 {
continue;
} else if i1 == i2 {
if len1 == len2 {
return true;
}
} else {
let mut len1 = len1;
let mut len2 = len2;
if c1 == 1 {
len1 -= 1;
}
if t1 == 0 {
len1 += 1;
}
if c2 == 1 {
len2 -= 1;
}
if t2 == 0 {
len2 += 1
}
if len1 == len2 {
return true;
}
}
}
}
false
}
}
题解 3 - rust
- 编辑时间：2023-01-08
- 执行用时：4ms
- 内存消耗：2.3MB
- 编程语言：rust
- 解法介绍：同上。
impl Solution {
pub fn is_it_possible(word1: String, word2: String) -> bool {
let mut list = [0; 26];
for i in 0..26 {
list[i] = i;
}
let mut m1 = [0; 26];
let mut m2 = [0; 26];
let mut len1 = 0;
let mut len2 = 0;
for c in word1.chars() {
let idx = c as usize - 'a' as usize;
m1[idx] += 1;
if m1[idx] == 1 {
len1 += 1;
}
}
for c in word2.chars() {
let idx = c as usize - 'a' as usize;
m2[idx] += 1;
if m2[idx] == 1 {
len2 += 1;
}
}
let check = |i1: usize, i2: usize| -> bool {
let c1 = m1[i1];
let t1 = m1[i2];
let c2 = m2[i2];
let t2 = m2[i1];
if c1 == 0 || c2 == 0 {
return false;
}
if i1 == i2 {
return len1 == len2;
}
if c1 == 1 && c2 == 1 {
if t1 == 0 && t2 == 0 {
return len1 == len2;
}
if t1 == 0 && t2 != 0 {
return len1 + 1 == len2;
}
if t1 != 0 && t2 == 0 {
return len1 == len2 + 1;
}
if t1 != 0 && t2 != 0 {
return len1 == len2;
}
}
if c1 == 1 && c2 > 1 {
if t1 == 0 && t2 == 0 {
return len1 == len2 + 1;
}
if t1 == 0 && t2 != 0 {
return len1 == len2;
}
if t1 != 0 && t2 == 0 {
return len1 - 1 == len2 + 1;
}
if t1 != 0 && t2 != 0 {
return len1 - 1 == len2;
}
}
if c1 > 1 && c2 == 1 {
if t1 == 0 && t2 == 0 {
return len2 == len1 + 1;
}
if t1 != 0 && t2 == 0 {
return len2 == len1;
}
if t1 == 0 && t2 != 0 {
return len2 - 1 == len1 + 1;
}
if t1 != 0 && t2 != 0 {
return len2 - 1 == len1;
}
}
if c1 > 1 && c2 > 1 {
if t1 == 0 && t2 == 0 {
return len1 == len2;
}
if t1 != 0 && t2 == 0 {
return len1 == len2 + 1;
}
if t1 == 0 && t2 != 0 {
return len1 + 1 == len2;
}
if t1 != 0 && t2 != 0 {
return len1 == len2;
}
}
false
};
for c1 in list.iter() {
for c2 in list.iter() {
if check(*c1, *c2) {
return true;
}
}
}
false
}
}