1202. 交换字符串中的元素

# 初始化，p存储每个点的父节点
p = list(range(n))

# 返回x的祖宗节点
def find(x):
    if p[x] != x:
        # 路径压缩
        p[x] = find(p[x])
    return p[x]


# 合并a和b所在的两个集合
p[find(a)] = find(b)

# 初始化，p存储每个点的父节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
p = list(range(n))
size = [1] * n

# 返回x的祖宗节点
def find(x):
    if p[x] != x:
        # 路径压缩
        p[x] = find(p[x])
    return p[x]

# 合并a和b所在的两个集合
if find(a) != find(b):
    size[find(b)] += size[find(a)]
    p[find(a)] = find(b)

# 初始化，p存储每个点的父节点，d[x]存储x到p[x]的距离
p = list(range(n))
d = [0] * n

# 返回x的祖宗节点
def find(x):
    if p[x] != x:
        t = find(p[x])
        d[x] += d[p[x]]
        p[x] = t
    return p[x]

# 合并a和b所在的两个集合
p[find(a)] = find(b)
d[find(a)] = distance

class Solution:
    def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
        def find(x):
            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)
        mp = defaultdict(list)
        for i, c in enumerate(s):
            heappush(mp[find(i)], c)
        return ''.join(heappop(mp[find(i)]) 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];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        for (List<Integer> pair : pairs) {
            p[find(pair.get(0))] = find(pair.get(1));
        }
        Map<Integer, PriorityQueue<Character>> mp = new HashMap<>();
        char[] chars = s.toCharArray();
        for (int i = 0; i < n; ++i) {
            mp.computeIfAbsent(find(i), k -> new PriorityQueue<>()).offer(chars[i]);
        }
        for (int i = 0; i < n; ++i) {
            chars[i] = mp.get(find(i)).poll();
        }
        return new String(chars);
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

class Solution {
public:
    vector<int> p;

    string smallestStringWithSwaps(string s, vector<vector<int>>& pairs) {
        int n = s.length();
        p.resize(n);
        for (int i = 0; i < n; ++i) p[i] = i;
        for (auto& pair : pairs) p[find(pair[0])] = find(pair[1]);
        unordered_map<int, vector<char>> mp;
        for (int i = 0; i < n; ++i) mp[find(i)].push_back(s[i]);
        for (auto& [k, v] : mp) sort(v.rbegin(), v.rend());
        string ans;
        for (int i = 0; i < n; ++i) {
            ans.push_back(mp[find(i)].back());
            mp[find(i)].pop_back();
        }
        return ans;
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }
};

func smallestStringWithSwaps(s string, pairs [][]int) string {
	n := len(s)
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	var find func(x int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	for _, pair := range pairs {
		p[find(pair[0])] = find(pair[1])
	}
	mp := make(map[int][]rune)
	for i, c := range s {
		mp[find(i)] = append(mp[find(i)], c)
	}
	for _, v := range mp {
		sort.Slice(v, func(i, j int) bool {
			return v[i] < v[j]
		})
	}
	var ans []rune
	for i := 0; i < n; i++ {
		ans = append(ans, mp[find(i)][0])
		mp[find(i)] = mp[find(i)][1:]
	}
	return string(ans)
}