---
comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/solution/1200-1299/1202.Smallest%20String%20With%20Swaps/README.md
rating: 1855
source: 第 155 场周赛 Q3
tags:
- 深度优先搜索
- 广度优先搜索
- 并查集
- 数组
- 哈希表
- 字符串
- 排序
---
# [1202. 交换字符串中的元素](https://leetcode.cn/problems/smallest-string-with-swaps)
[English Version](/solution/1200-1299/1202.Smallest%20String%20With%20Swaps/README_EN.md)
## 题目描述
给你一个字符串 s
,以及该字符串中的一些「索引对」数组 pairs
,其中 pairs[i] = [a, b]
表示字符串中的两个索引(编号从 0 开始)。
你可以 任意多次交换 在 pairs
中任意一对索引处的字符。
返回在经过若干次交换后,s
可以变成的按字典序最小的字符串。
示例 1:
输入:s = "dcab", pairs = [[0,3],[1,2]]
输出:"bacd"
解释:
交换 s[0] 和 s[3], s = "bcad"
交换 s[1] 和 s[2], s = "bacd"
示例 2:
输入:s = "dcab", pairs = [[0,3],[1,2],[0,2]]
输出:"abcd"
解释:
交换 s[0] 和 s[3], s = "bcad"
交换 s[0] 和 s[2], s = "acbd"
交换 s[1] 和 s[2], s = "abcd"
示例 3:
输入:s = "cba", pairs = [[0,1],[1,2]]
输出:"abc"
解释:
交换 s[0] 和 s[1], s = "bca"
交换 s[1] 和 s[2], s = "bac"
交换 s[0] 和 s[1], s = "abc"
提示:
1 <= s.length <= 10^5
0 <= pairs.length <= 10^5
0 <= pairs[i][0], pairs[i][1] < s.length
s
中只含有小写英文字母
## 解法
### 方法一:并查集
我们注意到,索引对具有传递性,即如果 $a$ 与 $b$ 可交换,而 $b$ 与 $c$ 可交换,那么 $a$ 与 $c$ 也可交换。因此,我们可以考虑使用并查集维护这些索引对的连通性,将属于同一个连通分量的字符按照字典序排序。
最后,遍历字符串,对于当前位置的字符,我们将其替换为该连通分量中最小的字符,然后从该连通分量中取出该字符,继续遍历字符串即可。
时间复杂度 $O(n \times \log n + m \times \alpha(m))$,空间复杂度 $O(n)$。其中 $n$ 和 $m$ 分别为字符串的长度和索引对的数量,而 $\alpha$ 为阿克曼函数的反函数。
#### Python3
```python
class Solution:
def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
def find(x: int) -> int:
if p[x] != x:
p[x] = find(p[x])
return p[x]
n = len(s)
p = list(range(n))
for a, b in pairs:
p[find(a)] = find(b)
d = defaultdict(list)
for i, c in enumerate(s):
d[find(i)].append(c)
for i in d.keys():
d[i].sort(reverse=True)
return "".join(d[find(i)].pop() for i in range(n))
```
#### Java
```java
class Solution {
private int[] p;
public String smallestStringWithSwaps(String s, List> pairs) {
int n = s.length();
p = new int[n];
List[] d = new List[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
d[i] = new ArrayList<>();
}
for (var pair : pairs) {
int a = pair.get(0), b = pair.get(1);
p[find(a)] = find(b);
}
char[] cs = s.toCharArray();
for (int i = 0; i < n; ++i) {
d[find(i)].add(cs[i]);
}
for (var e : d) {
e.sort((a, b) -> b - a);
}
for (int i = 0; i < n; ++i) {
var e = d[find(i)];
cs[i] = e.remove(e.size() - 1);
}
return String.valueOf(cs);
}
private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}
```
#### C++
```cpp
class Solution {
public:
string smallestStringWithSwaps(string s, vector>& pairs) {
int n = s.size();
int p[n];
iota(p, p + n, 0);
vector d[n];
function find = [&](int x) -> int {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
};
for (auto e : pairs) {
int a = e[0], b = e[1];
p[find(a)] = find(b);
}
for (int i = 0; i < n; ++i) {
d[find(i)].push_back(s[i]);
}
for (auto& e : d) {
sort(e.rbegin(), e.rend());
}
for (int i = 0; i < n; ++i) {
auto& e = d[find(i)];
s[i] = e.back();
e.pop_back();
}
return s;
}
};
```
#### Go
```go
func smallestStringWithSwaps(s string, pairs [][]int) string {
n := len(s)
p := make([]int, n)
d := make([][]byte, n)
for i := range p {
p[i] = i
}
var find func(int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
for _, pair := range pairs {
a, b := pair[0], pair[1]
p[find(a)] = find(b)
}
cs := []byte(s)
for i, c := range cs {
j := find(i)
d[j] = append(d[j], c)
}
for i := range d {
sort.Slice(d[i], func(a, b int) bool { return d[i][a] > d[i][b] })
}
for i := range cs {
j := find(i)
cs[i] = d[j][len(d[j])-1]
d[j] = d[j][:len(d[j])-1]
}
return string(cs)
}
```
#### TypeScript
```ts
function smallestStringWithSwaps(s: string, pairs: number[][]): string {
const n = s.length;
const p = new Array(n).fill(0).map((_, i) => i);
const find = (x: number): number => {
if (p[x] !== x) {
p[x] = find(p[x]);
}
return p[x];
};
const d: string[][] = new Array(n).fill(0).map(() => []);
for (const [a, b] of pairs) {
p[find(a)] = find(b);
}
for (let i = 0; i < n; ++i) {
d[find(i)].push(s[i]);
}
for (const e of d) {
e.sort((a, b) => b.charCodeAt(0) - a.charCodeAt(0));
}
const ans: string[] = [];
for (let i = 0; i < n; ++i) {
ans.push(d[find(i)].pop()!);
}
return ans.join('');
}
```
#### Rust
```rust
impl Solution {
#[allow(dead_code)]
pub fn smallest_string_with_swaps(s: String, pairs: Vec>) -> String {
let n = s.as_bytes().len();
let s = s.as_bytes();
let mut disjoint_set: Vec = vec![0; n];
let mut str_vec: Vec> = vec![Vec::new(); n];
let mut ret_str = String::new();
// Initialize the disjoint set
for i in 0..n {
disjoint_set[i] = i;
}
// Union the pairs
for pair in pairs {
Self::union(pair[0] as usize, pair[1] as usize, &mut disjoint_set);
}
// Initialize the return vector
for (i, c) in s.iter().enumerate() {
let p_c = Self::find(i, &mut disjoint_set);
str_vec[p_c].push(*c);
}
// Sort the return vector in reverse order
for cur_vec in &mut str_vec {
cur_vec.sort();
cur_vec.reverse();
}
// Construct the return string
for i in 0..n {
let index = Self::find(i, &mut disjoint_set);
ret_str.push(str_vec[index].last().unwrap().clone() as char);
str_vec[index].pop();
}
ret_str
}
#[allow(dead_code)]
fn find(x: usize, d_set: &mut Vec) -> usize {
if d_set[x] != x {
d_set[x] = Self::find(d_set[x], d_set);
}
d_set[x]
}
#[allow(dead_code)]
fn union(x: usize, y: usize, d_set: &mut Vec) {
let p_x = Self::find(x, d_set);
let p_y = Self::find(y, d_set);
d_set[p_x] = p_y;
}
}
```