| comments | difficulty | edit_url | rating | source | tags | |||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
true |
中等 |
1855 |
第 155 场周赛 Q3 |
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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 <= 10^50 <= pairs.length <= 10^50 <= pairs[i][0], pairs[i][1] < s.lengths中只含有小写英文字母
我们注意到，索引对具有传递性，即如果
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
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))class Solution {
private int[] p;
public String smallestStringWithSwaps(String s, List<List<Integer>> pairs) {
int n = s.length();
p = new int[n];
List<Character>[] 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];
}
}class Solution {
public:
string smallestStringWithSwaps(string s, vector<vector<int>>& pairs) {
int n = s.size();
int p[n];
iota(p, p + n, 0);
vector<char> d[n];
function<int(int)> 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;
}
};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)
}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('');
}impl Solution {
#[allow(dead_code)]
pub fn smallest_string_with_swaps(s: String, pairs: Vec<Vec<i32>>) -> String {
let n = s.as_bytes().len();
let s = s.as_bytes();
let mut disjoint_set: Vec<usize> = vec![0; n];
let mut str_vec: Vec<Vec<u8>> = 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>) -> 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<usize>) {
let p_x = Self::find(x, d_set);
let p_y = Self::find(y, d_set);
d_set[p_x] = p_y;
}
}