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LeetCode --- 450周赛

news 来源：原创 2025/5/26 5:51:29

题目列表

3550. 数位和等于下标的最小下标
3551. 数位和排序需要的最小交换次数
3552. 网格传送门旅游
3553. 包含给定路径的最小带权子树 II

一、数位和等于下标的最小下标

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直接模拟计算数位和即可，代码如下

// C++
class Solution {
public:int smallestIndex(vector<int>& nums) {int n = nums.size();for(int i = 0; i < n; i++){int x = nums[i], s = 0;while(x){s += x % 10;x /= 10;}if(s == i)return i;}return -1;}
};

# Python
class Solution:def smallestIndex(self, nums: List[int]) -> int:n = len(nums)for i, x in enumerate(nums):s = 0while x:s += x % 10x //= 10if i == s:return ireturn -1

二、数位和排序需要的最小交换次数

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由于我们可以通过排序得到每个数的最终位置，所以这题就转换成了给你两个元素相同的序列 A、B，要求将 A 通过交换转换成 B 的最小操作次数，通过对需要交换的位置加上有向边进行建图，问题就变成了对给定的有向图，如何使得它们各自成为一个独立的环的操作次数最少
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这里有一个结论：最少的交换次数为结点的个数减去环的个数

代码如下（关键就是计算环的个数）

// C++
class Union{
public:Union(int n):fa(n),cnt(n){iota(fa.begin(), fa.end(), 0);}void merge(int x, int y){int fa_x = find(x), fa_y = find(y);if(fa_x != fa_y){cnt--;fa[fa_x] = fa_y;}}int find(int x){int pa = x;while(pa != fa[pa]){pa = fa[pa];}while(x != pa){int tmp = fa[x];fa[x] = pa;x = tmp;}return pa;}int count(){return cnt;}
private:vector<int> fa;int cnt;
};
class Solution {
public:int minSwaps(vector<int>& nums) {int n = nums.size();vector<array<int, 3>> a(n);for (int i = 0; i < n; i++) {int s = 0;for (int x = nums[i]; x > 0; x /= 10) {s += x % 10;}a[i] = {s, nums[i], i};}ranges::sort(a);Union un(n);for(int i = 0; i < n; i++){un.merge(i, a[i][2]);}return n - un.count();}
};// 迭代
class Solution {
public:int minSwaps(vector<int>& nums) {int n = nums.size();vector<array<int, 3>> a(n);for (int i = 0; i < n; i++) {int s = 0;for (int x = nums[i]; x > 0; x /= 10) {s += x % 10;}a[i] = {s, nums[i], i};}ranges::sort(a);int c = 0;for(int i = 0; i < n; i++){if(a[i][2] < 0) continue;c++;int j = i;while(a[j][2] >= 0){int nxt = a[j][2];a[j][2] = -1;j = nxt;}}return n - c;}
};

三、网格传送门旅游

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很经典的最短路问题，可以用 Dijkstra算法 来解决，需要注意的每种传送门最多传送一次，重复传送显然是无意义的，A1->A2->A3 不如直接 A1->A3，代码如下

// C++
class Solution {static constexpr int dirs[4][2] = {{0,1}, {0,-1}, {1,0}, {-1,0}};
public:int minMoves(vector<string>& matrix) {int n = matrix.size(), m = matrix[0].size();vector<vector<pair<int,int>>> g(26);for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){if(isalpha(matrix[i][j])){g[matrix[i][j] - 'A'].emplace_back(i, j);}}}priority_queue<tuple<int,int,int>, vector<tuple<int,int,int>>, greater<>> pq;pq.emplace(0, 0, 0);vector dist(n, vector<int>(m, INT_MAX));dist[0][0] = 0;while(pq.size()){auto [d, x, y] = pq.top(); pq.pop();if(x == n - 1 && y == m - 1) return d;if(d != dist[x][y]) continue;if(isalpha(matrix[x][y])){int i = matrix[x][y] - 'A';for(auto [xi, yi] : g[i]){if(dist[xi][yi] > d){dist[xi][yi] = d;pq.emplace(d, xi, yi);}}g[i].resize(0);}for(int i = 0; i < 4; i++){int xi = x + dirs[i][0], yi = y + dirs[i][1];if(xi < 0 || xi >= n || yi < 0 || yi >= m || matrix[xi][yi] == '#' || dist[xi][yi] <= d + 1)continue;dist[xi][yi] = d + 1;pq.emplace(d + 1, xi, yi);}}return -1;}
};// 0-1 bfs
// 由于本题的距离只有 +0 / +1 两种选项，所以这里可以直接用双端队列模拟优先级队列（堆）
// +0 的距离依旧是队列中最小的距离，所以头插
// +1 的距离会成为最大值，所以尾插，如果不理解，建立手动模拟一下
class Solution {static constexpr int dirs[4][2] = {{0,1}, {0,-1}, {1,0}, {-1,0}};
public:int minMoves(vector<string>& matrix) {int n = matrix.size(), m = matrix[0].size();vector<vector<pair<int,int>>> g(26);for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){if(isalpha(matrix[i][j])){g[matrix[i][j] - 'A'].emplace_back(i, j);}}}deque<tuple<int,int,int>> dq;dq.push_back({0, 0, 0});vector dist(n, vector<int>(m, INT_MAX));dist[0][0] = 0;while(dq.size()){auto [d, x, y] = dq.front(); dq.pop_front();if(x == n - 1 && y == m - 1) return d;if(d != dist[x][y]) continue;if(isalpha(matrix[x][y])){int i = matrix[x][y] - 'A';for(auto [xi, yi] : g[i]){if(dist[xi][yi] > d){dist[xi][yi] = d;dq.push_front({d, xi, yi});}}g[i].resize(0);}for(int i = 0; i < 4; i++){int xi = x + dirs[i][0], yi = y + dirs[i][1];if(xi < 0 || xi >= n || yi < 0 || yi >= m || matrix[xi][yi] == '#' || dist[xi][yi] <= d + 1)continue;dist[xi][yi] = d + 1;dq.push_back({d + 1, xi, yi});}}return -1;}
};

四、包含给定路径的最小带权子树 II

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本题要求查询三个结点能够互相到达的最小权重。
思路：

两个结点之间能互相到达的最小权重，即两个结点的最短路径，如何求解？假设这两个结点为 A、B，它们的最近公共祖先为 fa，则 dist(A,B) = dist(A,root) + dist(B,root) - 2*dist(fa, root)，其中 dist(x,y) 表示 x 到 y 的距离，而我们很容易得出每个结点到根节点的距离
如果是三个结点呢？

显然三点之间的最小权重等于 dist(A,B,C) = [dist(A,B) + dist(B,C) + dist(C, A)] / 2
如何求解两个节点的最近公共祖先（LCA）？用树上倍增算法，能在 O(logn) 的时间内计算出来

代码如下

// C++
class LcaBinaryLifting{vector<int> dist, depth;vector<vector<int>> pa;
public:LcaBinaryLifting(vector<vector<int>>& edges){int n = edges.size() + 1;int m = bit_width((unsigned) n);vector<vector<pair<int,int>>> g(n);for(auto & e : edges){g[e[0]].emplace_back(e[1], e[2]);g[e[1]].emplace_back(e[0], e[2]);}depth.resize(n);dist.resize(n);pa.resize(n, vector<int>(m, -1));auto dfs = [&](this auto && dfs, int x, int fa, int s, int d)->void{dist[x] = s;pa[x][0] = fa;depth[x] = d;for(auto [y, w] : g[x]){if(y != fa){dfs(y, x, w + s, d + 1);}}};dfs(0, -1, 0, 0);for(int j = 1; j < m; j++){for(int i = 0; i < n; i++){if(pa[i][j-1] != -1){pa[i][j] = pa[pa[i][j-1]][j-1];}}}}int getLCA(int a, int b){if(depth[a] < depth[b]){swap(a, b);}for(int k = depth[a] - depth[b]; k; k &= k - 1){a = pa[a][countr_zero((unsigned)k)];}if(a == b) return a;// 处于同一个深度for(int j = pa[a].size() - 1; j >= 0; j--){if(pa[a][j] != pa[b][j]){a = pa[a][j];b = pa[b][j];}}return pa[a][0];}int getDist(int a, int b){int c = getLCA(a, b);return dist[a] + dist[b] - dist[c] * 2;}
};
class Solution {
public:vector<int> minimumWeight(vector<vector<int>>& edges, vector<vector<int>>& queries) {LcaBinaryLifting g(edges);vector<int> ans(queries.size());for(int i = 0; i < queries.size(); i++){int a = queries[i][0], b = queries[i][1], c = queries[i][2];ans[i] = (g.getDist(a, b) + g.getDist(b, c) + g.getDist(c, a)) / 2;}return ans;}
};

# Python
class LcaBinaryLifting:def __init__(self, edges: List[List[int]]):n = len(edges) + 1m = n.bit_length()g = [[] for _ in range(n)]for x, y, w in edges:g[x].append((y, w))g[y].append((x, w))depth = [0] * ndis = [0] * npa = [[-1] * m for _ in range(n)]def dfs(x:int, fa:int)->int:pa[x][0] = fafor y, w in g[x]:if y != fa:depth[y] = depth[x] + 1dis[y] = dis[x] + wdfs(y, x)dfs(0, -1)for j in range(m - 1):for i in range(n):if pa[i][j] != -1:pa[i][j+1] = pa[pa[i][j]][j]self.depth = depthself.dis = disself.pa = padef get_kth_ancestor(self, node:int, k:int)->int:for i in range(k.bit_length()):if k >> i & 1:node = self.pa[node][i]return nodedef get_lca(self, x:int, y:int)->int:if self.depth[x] > self.depth[y]:x, y = y, xy = self.get_kth_ancestor(y, self.depth[y] - self.depth[x])if x == y:return xfor i in range(len(self.pa[x]) - 1, -1, -1):px, py = self.pa[x][i], self.pa[y][i]if px != py:x, y = py, pxreturn self.pa[x][0]def get_dis(self, x:int, y:int)->int:return self.dis[x] + self.dis[y] - 2 * self.dis[self.get_lca(x, y)]
class Solution:def minimumWeight(self, edges: List[List[int]], queries: List[List[int]]) -> List[int]:g = LcaBinaryLifting(edges)ans = []for a, b, c in queries:ans.append((g.get_dis(a, b) + g.get_dis(a, c) + g.get_dis(b, c))//2)return ans