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中等
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第 155 场周赛 Q3
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

English Version

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$$b$ 可交换,而 $b$$c$ 可交换,那么 $a$$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$$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

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

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];
    }
}

C++

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;
    }
};

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

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

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;
    }
}