November 2017 – Page 2 – Huahua’s Tech Road

花花酱 LeetCode 113. Path Sum II

By zxi on November 27, 2017

Problem:

Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

/ \

4 8

/ / \

11 13 4

/ \ / \

7 2 5 1

return

[

[5,4,11,2],

[5,8,4,5]

]

Idea:

Recursion

Solution:

C++

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

vector<vector<int>> pathSum(TreeNode* root, int sum) {

vector<vector<int>> ans;

vector<int> curr;

pathSum(root, sum, curr, ans);

return ans;

}

private:

void pathSum(TreeNode* root, int sum, vector<int>& curr, vector<vector<int>>& ans) {

if(root==nullptr) return;

if(root->left==nullptr && root->right==nullptr) {

if(root->val == sum) {

ans.push_back(curr);

ans.back().push_back(root->val);

}

return;

}

curr.push_back(root->val);

int new_sum = sum - root->val;

pathSum(root->left, new_sum, curr, ans);

pathSum(root->right, new_sum, curr, ans);

curr.pop_back();

}

};

Related Problems:

花花酱 LeetCode 53. Maximum Subarray

By zxi on November 27, 2017

Problem:

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.

Idea:

Solution:

C++

// Author: Huahua

// Runtime: 6 ms (better than 98.66%)

class Solution {

public:

int maxSubArray(vector<int>& nums) {

vector<int> f(nums.size());

f[0] = nums[0];

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

f[i] = max(f[i - 1] + nums[i], nums[i]);

return *std::max_element(f.begin(), f.end());

}

};

花花酱 LeetCode 657. Judge Route Circle

By zxi on November 27, 2017

Problem:

Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place.

The move sequence is represented by a string. And each move is represent by a character. The valid robot moves are R (Right), L (Left), U (Up) and D (down). The output should be true or false representing whether the robot makes a circle.

Example 1:

1 2	Input: "UD" Output: true

Example 2:

1 2	Input: "LL" Output: false

Idea:

Simulation

Solution:

C++

// Author: Huahua

// Runtime: 19 ms

class Solution {

public:

bool judgeCircle(string moves) {

int cx{0};

int cy{0};

for (const char move : moves) {

switch(move) {

case 'U':

cy--;

break;

case 'D':

cy++;

break;

case 'R':

cx++;

break;

case 'L':

cx--;

break;

}

return cx==0 && cy==0;

}

};

花花酱 LeetCode 312. Burst Balloons

By zxi on November 27, 2017

Problem:

Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and rightthen becomes adjacent.

Find the maximum coins you can collect by bursting the balloons wisely.

Note:
(1) You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
(2) 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Given [3, 1, 5, 8]

Return 167

1 2	nums = [3,1,5,8] --> [3,5,8] --> [3,8] --> [8] --> [] coins = 315 + 358 + 138 + 181 = 167

Idea

Solution1:

C++ / Recursion with memoization

// Author: Huahua

// Runtime: 16 ms

class Solution {

public:

int maxCoins(vector<int>& nums) {

int n = nums.size();

nums.insert(nums.begin(), 1);

nums.push_back(1);

// c[i][j] = maxCoins(nums[i:j+1])

c_ = vector<vector<int>>(n+2, vector<int>(n+2, 0));

return maxCoins(nums, 1, n);

}

private:

int maxCoins(const vector<int>& nums, const int i, const int j) {

if(i>j) return 0;

if(c_[i][j]>0) return c_[i][j];

if(i==j) return nums[i-1]*nums[i]*nums[i+1];

int ans=0;

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

ans=max(ans, maxCoins(nums,i,k-1) + nums[i-1]*nums[k]*nums[j+1] + maxCoins(nums, k+1, j));

c_[i][j] = ans;

return c_[i][j];

}

vector<vector<int>> c_;

};

Java

// Author: Huahua

class Solution {

private int[][] memo;

private int[] vals;

public int maxCoins(int[] nums) {

final int n = nums.length;

vals = new int[n + 2];

for (int i = 0; i < n; ++i) vals[i + 1] = nums[i];

vals[0] = vals[n + 1] = 1;

memo = new int[n + 2][n + 2];

return maxCoins(1, n);

}

private int maxCoins(int i, int j) {

if (i > j) return 0;

if (i == j) return vals[i - 1] * vals[i] * vals[i + 1];

if (memo[i][j] > 0) return memo[i][j];

int ans = 0;

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

ans = Math.max(ans, maxCoins(i, k - 1) + vals[i - 1] * vals[k] * vals[j + 1] + maxCoins(k + 1, j));

memo[i][j] = ans;

return memo[i][j];

}

Solution2:

C++ / DP

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

int maxCoins(vector<int>& nums) {

int n = nums.size();

nums.insert(nums.begin(), 1);

nums.push_back(1);

// c[i][j] = maxCoins(nums[i:j+1])

vector<vector<int>> c(n+2, vector<int>(n+2, 0));

for(int l=1;l<=n;++l)

for(int i=1;i<=n-l+1;++i) {

int j=i+l-1;

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

c[i][j] = max(c[i][j], c[i][k-1] + nums[i-1]*nums[k]*nums[j+1] + c[k+1][j]);

}

return c[1][n];

}

};

Java / DP

Java

// Author: Huahua

class Solution {

public int maxCoins(int[] nums) {

final int n = nums.length;

int[] vals = new int[n + 2];

for (int i = 0; i < n; ++i) vals[i + 1] = nums[i];

vals[0] = vals[n + 1] = 1;

int[][] dp = new int[n + 2][n + 2];

for (int l = 1; l <= n; ++l)

for (int i = 1; i + l <= n + 1; ++i) {

int j = i + l - 1;

int best = 0;

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

best = Math.max(best,

dp[i][k - 1] + vals[i - 1] * vals[k] * vals[j + 1] + dp[k + 1][j]);

dp[i][j] = best;

}

return dp[1][n];

}

花花酱 LeetCode 1. Two Sum

By zxi on November 23, 2017

Problem:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Idea:

Brute force / Hashtable

Solution1:

Brute force / C++

Time complexity: O(n^2)

Space complexity: O(1)

// Author: Huahua

// Runtime: 166 ms

class Solution {

public:

vector<int> twoSum(vector<int>& nums, int target) {

for (int i = 0; i < nums.size(); ++i) {

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

int sum = nums[i] + nums[j];

if (sum == target) {

return {i, j};

}

return {};

}

};

Solution 2:

Hashtable / C++

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

vector<int> twoSum(vector<int>& nums, int target) {

unordered_map<int, int> indies;

for (int i = 0; i < nums.size(); ++i)

indies[nums[i]] = i;

for (int i = 0; i < nums.size(); ++i) {

int left = target - nums[i];

if (indies.count(left) && indies[left] != i) {

return {i, indies[left]};

}

return {};

}

};

Posts published in November 2017

花花酱 LeetCode 113. Path Sum II

花花酱 LeetCode 53. Maximum Subarray

花花酱 LeetCode 657. Judge Route Circle

花花酱 LeetCode 312. Burst Balloons

Java

Java

花花酱 LeetCode 1. Two Sum