head – Huahua’s Tech Road

花花酱 LeetCode 514. Freedom Trail

By zxi on April 6, 2025

一眼DP，看数据规模标程应该是O(100³)。

阶段肯定就是吟唱完一个字符，状态和哪个字符在12:00点有关。

所以我们定义 dp[i][j] := 吟唱完keys[0~i]后，rings[j]停在12:00点所需要的最少步数。

转移方程：dp[i][j] = min(dp[i-1][k] + steps(k, j) + 1) if keys[i] == rings[j] else +inf.

第i-1字符吟唱完ring停留在k，第i个字符吟唱完ring停留在j (rings[j] 必须和 keys[i] 相同)，steps(k, j)表示ring从k转到j所需要的最少步数，最后还要+1，表示吟唱当前字符所需要的额外步骤。

边界条件：对于第0个字符，ring一开始停留在第0个字符，最小步数就是1 + steps(0, key[0])。

时间复杂度：O(mn²)
空间复杂度：O(mn) 可以降维到 O(n)

class Solution {

public:

int findRotateSteps(string ring, string key) {

const int n = ring.size();

const int m = key.size();

auto steps = [n](int i, int j) {

return min(abs(i - j), n - abs(i - j));

};

// dp[i][j] min steps to spell key[0~i] and ring[j] is at 12:00.

vector<vector<int>> dp(m, vector<int>(n, INT_MAX/2));

for (int i = 0; i < m; ++i)

for (int j = 0; j < n; ++j) {

if (ring[j] != key[i]) continue;

for (int k = 0; k < n; ++k)

dp[i][j] = min(dp[i][j], 1 + (i == 0 ? steps(0, j) : (dp[i - 1][k] + steps(j, k))));

}

return *min_element(begin(dp.back()), end(dp.back()));

}

};

花花酱 LeetCode 778. Swim in Rising Water

By zxi on January 31, 2019

On an N x N grid, each square grid[i][j] represents the elevation at that point (i,j).

Now rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.

You start at the top left square (0, 0). What is the least time until you can reach the bottom right square (N-1, N-1)?

Example 1:

Input: [[0,2],[1,3]]
Output: 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.

You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.

Example 2:

Input: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
Output: 16
Explanation:
 0  1  2  3  4
24 23 22 21  5
12 13 14 15 16
11 17 18 19 20
10  9  8  7  6

The final route is marked in bold.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.

Note:

2 <= N <= 50.
grid[i][j] is a permutation of [0, …, N*N – 1].

Solution 1: Dijkstra’s Algorithm

Time complexity: O(n^2*logn)
Space complexity: O(n^2)

C++

// Author: Huahua, running time: 8 ms

class Solution {

public:

int swimInWater(vector<vector<int>>& grid) {

const int n = grid.size();

priority_queue<pair<int, int>> q; // {-time, y * N + x}

q.push({-grid[0][0], 0 * n + 0});

vector<int> seen(n * n);

vector<int> dirs{-1, 0, 1, 0, -1};

seen[0 * n + 0] = 1;

while (!q.empty()) {

auto node = q.top(); q.pop();

int t = -node.first;

int x = node.second % n;

int y = node.second / n;

if (x == n - 1 && y == n - 1) return t;

for (int i = 0; i < 4; ++i) {

int tx = x + dirs[i];

int ty = y + dirs[i + 1];

if (tx < 0 || ty < 0 || tx >= n || ty >= n) continue;

if (seen[ty * n + tx]) continue;

seen[ty * n + tx] = 1;

q.push({-max(t, grid[ty][tx]), ty * n + tx});

}

return -1;

}

};

Solution 2: Binary Search + BFS

Time complexity: O(2logn * n^2)
Space complexity: O(n^2)

C++

// Author: Huahua, running time: 8 ms

class Solution {

public:

int swimInWater(vector<vector<int>>& grid) {

const int n = grid.size();

auto hasPath = [&grid, n](int t) {

if (grid[0][0] > t) return false;

queue<int> q;

vector<int> seen(n * n);

vector<int> dirs{1, 0, -1, 0, 1};

q.push(0);

while (!q.empty()) {

const int x = q.front() % n;

const int y = q.front() / n;

q.pop();

if (x == n - 1 && y == n - 1) return true;

for (int i = 0; i < 4; ++i) {

const int tx = x + dirs[i];

const int ty = y + dirs[i + 1];

if (tx < 0 || ty < 0 || tx >= n || ty >= n || grid[ty][tx] > t) continue;

const int key = ty * n + tx;

if (seen[key]) continue;

seen[key] = 1;

q.push(key);

}

return false;

};

int l = 0;

int r = n * n;

while (l < r) {

int m = l + (r - l) / 2;

if (hasPath(m)) {

r = m;

} else {

l = m + 1;

}

return l;

}

};

花花酱 LeetCode 725. Split Linked List in Parts

By zxi on November 13, 2017

Problem:

Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list “parts”.

The length of each part should be as equal as possible: no two parts should have a size differing by more than 1. This may lead to some parts being null.

The parts should be in order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal parts occurring later.

Return a List of ListNode’s representing the linked list parts that are formed.

Examples 1->2->3->4, k = 5 // 5 equal parts [ [1], [2], [3], [4], null ]

Example 1:

Input:

root = [1, 2, 3], k = 5

Output: [[1],[2],[3],[],[]]

Explanation:

The input and each element of the output are ListNodes, not arrays.

For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3, and root.next.next.next = null.

The first element output[0] has output[0].val = 1, output[0].next = null.

The last element output[4] is null, but it's string representation as a ListNode is [].

Example 2:

Input:

root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3

Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]

Explanation:

The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.

Note:

The length of root will be in the range [0, 1000].
Each value of a node in the input will be an integer in the range [0, 999].
k will be an integer in the range [1, 50].

Idea:

List + Simulation

Solution:

C++

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

vector<ListNode*> splitListToParts(ListNode* root, int k) {

int len = 0;

for (ListNode* head = root; head; head = head->next) ++len;

vector<ListNode*> ans(k, nullptr);

int l = len / k;

int r = len % k;

ListNode* head = root;

ListNode* prev = nullptr;

for (int i = 0; i < k; ++i, --r) {

ans[i] = head;

for (int j = 0; j < l + (r > 0); ++j) {

prev = head;

head = head->next;

}

if (prev) prev->next = nullptr;

}

return ans;

}

};

Java

// Author: Huahua

// Runtime: 4 ms

class Solution {

public ListNode[] splitListToParts(ListNode root, int k) {

ListNode[] ans = new ListNode[k];

int len = 0;

for (ListNode head = root; head != null; head = head.next) ++len;

int l = len / k;

int r = len % k;

ListNode prev = null;

ListNode head = root;

for (int i = 0; i < k; ++i, --r) {

ans[i] = head;

int part_len = l + ((r > 0) ? 1 : 0);

for (int j = 0; j < part_len; ++j) {

prev = head;

head = head.next;

}

if (prev != null) prev.next = null;

}

return ans;

}

Python

"""

Author: Huahua

Runtime: 79 ms

"""

class Solution:

def splitListToParts(self, root, k):

total_len = 0

head = root

while head:

total_len += 1

head = head.next

ans = [None for _ in range(k)]

l, r = total_len // k, total_len % k

prev, head = None, root

for i in range(k):

ans[i] = head

for j in range(l + (1 if r > 0 else 0)):

prev, head = head, head.next

if prev: prev.next = None

r -= 1

return ans

Posts tagged as “head”

花花酱 LeetCode 514. Freedom Trail

花花酱 LeetCode 778. Swim in Rising Water

Solution 1: Dijkstra’s Algorithm

C++

Solution 2: Binary Search + BFS

C++

花花酱 LeetCode 725. Split Linked List in Parts