Huahua’s Tech Road

花花酱 LeetCode 225. Implement Stack using Queues

By zxi on March 13, 2018

题目大意：用队列来实现栈。

Problem:

https://leetcode.com/problems/implement-stack-using-queues/description/

Implement the following operations of a stack using queues.

push(x) — Push element x onto stack.
pop() — Removes the element on top of the stack.
top() — Get the top element.
empty() — Return whether the stack is empty.

Notes:

You must use only standard operations of a queue — which means only push to back, peek/pop from front, size, and is empty operations are valid.
Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

Idea:

Using a single queue, for every push, shift the queue (n – 1) times such that the last element becomes the first element in the queue.

e.g.

push(1): q: [1]

push(2): q: [1, 2] -> [2, 1]

push(3): q: [2, 1, 3] -> [1, 3, 2] -> [3, 2, 1]

push(4): q: [3, 2, 1, 4] -> [2, 1, 4, 3] -> [1, 4, 3, 2] -> [4, 3, 2, 1]

Solution:

Time complexity:

Push: O(n)

Pop/top/empty: O(1)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 3 ms

class MyStack {

public:

/** Initialize your data structure here. */

MyStack() {}

/** Push element x onto stack. */

void push(int x) {

q_.push(x);

for (int i = 1; i < q_.size(); ++i) {

q_.push(q_.front());

q_.pop();

}

/** Removes the element on top of the stack and returns that element. */

int pop() {

int top = q_.front();

q_.pop();

return top;

}

/** Get the top element. */

int top() {

return q_.front();

}

/** Returns whether the stack is empty. */

bool empty() {

return q_.empty();

}

private:

queue<int> q_;

};

花花酱 LeetCode 799. Champagne Tower

By zxi on March 11, 2018

题目大意：往一个香槟塔（i层有i个杯子）倒入n个杯子容量的香槟之后，求指定位置杯子中酒的容量。

Problem:

https://leetcode.com/problems/champagne-tower/description/

We stack glasses in a pyramid, where the first row has 1 glass, the second row has 2 glasses, and so on until the 100th row. Each glass holds one cup (250ml) of champagne.

Then, some champagne is poured in the first glass at the top. When the top most glass is full, any excess liquid poured will fall equally to the glass immediately to the left and right of it. When those glasses become full, any excess champagne will fall equally to the left and right of those glasses, and so on. (A glass at the bottom row has it’s excess champagne fall on the floor.)

For example, after one cup of champagne is poured, the top most glass is full. After two cups of champagne are poured, the two glasses on the second row are half full. After three cups of champagne are poured, those two cups become full – there are 3 full glasses total now. After four cups of champagne are poured, the third row has the middle glass half full, and the two outside glasses are a quarter full, as pictured below.

Now after pouring some non-negative integer cups of champagne, return how full the j-th glass in the i-th row is (both i and j are 0 indexed.)

Example 1:

Input: poured = 1, query_glass = 1, query_row = 1

Output: 0.0

Explanation: We poured 1 cup of champagne to the top glass of the tower (which is indexed as (0, 0)). There will be no excess liquid so all the glasses under the top glass will remain empty.

Example 2:

Input: poured = 2, query_glass = 1, query_row = 1

Output: 0.5

Explanation: We poured 2 cups of champagne to the top glass of the tower (which is indexed as (0, 0)). There is one cup of excess liquid. The glass indexed as (1, 0) and the glass indexed as (1, 1) will share the excess liquid equally, and each will get half cup of champange.

Note:

poured will be in the range of [0, 10 ^ 9].
query_glass and query_row will be in the range of [0, 99].

Idea: DP + simulation

define dp[i][j] as the volume of champagne will be poured into the j -th glass in the i-th row, dp[i][j] can be greater than 1.

Init dp[0][0] = poured.

Transition: if dp[i][j] > 1, it will overflow, the overflow part will be evenly distributed to dp[i+1][j], dp[i+1][j+1]

if dp[i][j] > 1:
dp[i+1][j] += (dp[i][j] – 1) / 2.0
dp[i+1][j + 1] += (dp[i][j] – 1) / 2.0

Answer: min(1.0, dp[query_row][query_index])

Solution 1:

C++

Time complexity: O(100*100)

Space complexity: O(100*100)

// Author: Huahua

// Running time: 22 ms

class Solution {

public:

double champagneTower(int poured, int query_row, int query_glass) {

constexpr int kRows = 100;

vector<vector<double>> dp(kRows, vector<double>(kRows));

dp[0][0] = poured;

for (int i = 0; i < kRows - 1; ++i)

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

if (dp[i][j] > 1) {

dp[i + 1][j ] += (dp[i][j] - 1) / 2.0;

dp[i + 1][j + 1] += (dp[i][j] - 1) / 2.0;

dp[i][j] = 1.0;

}

return std::min(1.0, dp[query_row][query_glass]);

}

};

Pull

// Author: Huahua

// Running time: 27 ms

class Solution {

public:

double champagneTower(int poured, int query_row, int query_glass) {

constexpr int kRows = 100;

vector<vector<double>> dp(kRows, vector<double>(kRows));

dp[0][0] = poured;

for (int i = 1; i < kRows; ++i)

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

if (j > 0) dp[i][j] += max(0.0, (dp[i - 1][j - 1] - 1) / 2.0);

if (j < i) dp[i][j] += max(0.0, (dp[i - 1][j ] - 1) / 2.0);

}

return std::min(1.0, dp[query_row][query_glass]);

}

};

C++

Time complexity: O(rows * 100)

Space complexity: O(100)

Push

// Author: Huahua

// Running time: 16 ms

class Solution {

public:

double champagneTower(int poured, int query_row, int query_glass) {

constexpr int kCols = 100;

vector<double> dp(kCols);

dp[0] = poured;

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

vector<double> tmp(kCols);

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

if (dp[j] > 1) {

tmp[j ] += (dp[j] - 1) / 2.0;

tmp[j + 1] += (dp[j] - 1) / 2.0;

}

dp.swap(tmp);

}

return std::min(1.0, dp[query_glass]);

}

};

Pull

class Solution {

public:

double champagneTower(int poured, int query_row, int query_glass) {

constexpr int kCols = 100;

vector<double> dp(kCols);

dp[0] = poured;

for (int i = 1; i <= query_row; ++i) {

vector<double> tmp(kCols);

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

if (j > 0) tmp[j] += max(0.0, (dp[j - 1] - 1) / 2.0);

if (j < i) tmp[j] += max(0.0, (dp[j ] - 1) / 2.0);

}

dp.swap(tmp);

}

return std::min(1.0, dp[query_glass]);

}

};

花花酱 LeetCode 797. All Paths From Source to Target

By zxi on March 10, 2018

题目大意：给你一个无环有向图，返回所有从节点0到节点n-1的路径。

Problem:

https://leetcode.com/problems/all-paths-from-source-to-target/description

Given a directed, acyclic graph of N nodes. Find all possible paths from node 0 to node N-1, and return them in any order.

The graph is given as follows: the nodes are 0, 1, …, graph.length – 1. graph[i] is a list of all nodes j for which the edge (i, j) exists.

Example:

Input: [[1,2], [3], [3], []]

Output: [[0,1,3],[0,2,3]]

Explanation: The graph looks like this:

0--->1

| |

v v

2--->3

There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.

Note:

The number of nodes in the graph will be in the range [2, 15].
You can print different paths in any order, but you should keep the order of nodes inside one path.

Solution 1: DFS

Time complexity: O(n!)

Space complexity: O(n)

// Author: Huahua

// Running time: 85 ms

class Solution {

public:

vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {

vector<vector<int>> ans;

vector<int> path(1, 0);

dfs(graph, 0, graph.size() - 1, path, ans);

return ans;

}

private:

void dfs(const vector<vector<int>>& graph,

int cur, int dest, vector<int>& path, vector<vector<int>>& ans) {

if (cur == dest) {

ans.push_back(path);

return;

}

for (int next : graph[cur]) {

path.push_back(next);

dfs(graph, next, dest, path, ans);

path.pop_back();

}

};

“Cleaner” Version

class Solution {

public:

vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {

vector<vector<int>> ans;

vector<int> path{0};

dfs(graph, path, ans);

return ans;

}

private:

void dfs(const vector<vector<int>>& graph,

vector<int>& path, vector<vector<int>>& ans) {

if (path.back() == graph.size() - 1) {

ans.push_back(path);

return;

}

for (int next : graph[path.back()]) {

path.push_back(next);

dfs(graph, path, ans);

path.pop_back();

}

};

Huahua's Tech Road

花花酱 LeetCode 203. Remove Linked List Elements

花花酱 LeetCode 225. Implement Stack using Queues

花花酱 LeetCode 234. Palindrome Linked List

花花酱 LeetCode 799. Champagne Tower

花花酱 LeetCode 797. All Paths From Source to Target