Implement pow(x, n).
class Solution {
double power(double x, long long n) {
if (n < 0) return 1 / power(x, -n);
if (n == 0) return 1;
double t = power(x, n >> 1);
if ((n&1) == 0)
return t * t;
else
return t * t * x;
}
public:
double pow(double x, int n) {
return power(x, (long long)n);
}
};Given a number represented as an array of digits, plus one to the number.
class Solution {
public:
vector<int> plusOne(vector<int> &digits) {
reverse(digits.begin(), digits.end());
digits[0] += 1;
int i = 0;
while (i < digits.size() && digits[i] > 9) {
digits[i] = 0;
if (i == digits.size() - 1)
digits.push_back(1);
else
digits[i + 1] += 1;
++i;
}
reverse(digits.begin(), digits.end());
return digits;
}
};Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.
For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *partition(ListNode *head, int x) {
ListNode *root1 = new ListNode(0), *root2 = new ListNode(0);
ListNode *p = head, *q1 = root1, *q2 = root2;
while (p != NULL) {
if (p->val < x)
q1->next = p, q1 = p;
else
q2->next = p, q2 = p;
p = p->next;
}
q1->next = root2->next;
q2->next = NULL;
p = root1->next;
delete root1;
delete root2;
return p;
}
};Determine whether an integer is a palindrome. Do this without extra space.
class Solution {
public:
bool isPalindrome(int x) {
if (x < 0) return false;
int rev = 0, t = x;
while (t)
rev = rev * 10 + t % 10, t /= 10;
return (rev == x);
}
};The set [1,2,3,…,n] contains
a total of n! unique permutations.
By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):
"123""132""213""231""312""321"
Given n and k, return the kth permutation sequence.
Note: Given n will be between 1 and 9 inclusive.
class Solution {
public:
string getPermutation(int n, int k) {
string ans;
--k;
vector<bool> f(n, false);
long long s = 1;
for (int i = 1; i < n; ++i) s *= i;
for (int i = 0; i < n; ++i) {
int t = k / s;
int j = 0, c = 0;
while (c <= t) {
if (!f[j]) ++c;
++j;
}
f[j - 1] = true;
ans += ('0' + j);
k %= s;
if (i != n - 1) s /= (n - 1 - i);
}
return ans;
}
};Given numRows, generate the first numRows of Pascal's triangle.
For example, given numRows = 5,
Return
[
[1],
[1,1],
[1,2,1],
[1,3,3,1],
[1,4,6,4,1]
]
class Solution {
public:
vector<vector<int> > generate(int numRows) {
vector<vector<int> > res;
for (int n = 0; n < numRows; ++n) {
vector<int> a(n + 1, 1);
for (int m = 1; m <= n; ++m)
a[m] = a[m - 1] * (n - m + 1) / m;
res.push_back(a);
}
return res;
}
};Given an index k, return the kth row of the Pascal's triangle.
For example, given k = 3,
Return [1,3,3,1].
Note:
Could you optimize your algorithm to use only O(k) extra space?
class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<int> res(rowIndex + 1, 1);
for (int m = 1; m <= rowIndex; ++m)
res[m] = (long long)res[m - 1] * (rowIndex - m + 1) / m;
return res;
}
};Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:Given the below binary tree and
sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which
sum is 22.
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool hasPathSum(TreeNode *root, int sum) {
if (root == NULL) return false;
if (root->left == NULL && root->right == NULL)
return (root->val == sum);
return (hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val));
}
};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,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
vector<vector<int> > res;
list<int> lst;
void dfs(TreeNode *root, int sum) {
if (root->left == NULL && root->right == NULL) {
lst.push_back(root->val);
if (root->val == sum) res.push_back(vector<int>(lst.begin(), lst.end()));
lst.pop_back();
return;
}
lst.push_back(root->val);
if (root->left != NULL)
dfs(root->left, sum - root->val);
if (root->right != NULL)
dfs(root->right, sum - root->val);
lst.pop_back();
}
public:
vector<vector<int> > pathSum(TreeNode *root, int sum) {
if (root != NULL) {
dfs(root, sum);
}
return res;
}
};Populating Next Right Pointers in Each Node
Total Accepted: 9862 Total Submissions: 28977My SubmissionsGiven a binary tree
struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
}
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
Initially, all next pointers are set to NULL.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1
/ \
2 3
/ \ / \
4 5 6 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ / \
4->5->6->7 -> NULL
/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
* int val;
* TreeLinkNode *left, *right, *next;
* TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
if (root == NULL) return;
queue<TreeLinkNode*> q;
q.push(root);
int cnt = 1;
while (!q.empty()) {
TreeLinkNode *fa = NULL;
int n = cnt;
cnt = 0;
while (n--) {
TreeLinkNode *p = q.front();
q.pop();
if (p->right != NULL) {
q.push(p->right);
++cnt;
}
if (p->left != NULL) {
q.push(p->left);
++cnt;
}
p->next = fa;
fa = p;
}
}
}
};Populating Next Right Pointers in Each Node II
Total Accepted: 6907 Total Submissions: 23625My SubmissionsFollow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
- You may only use constant extra space.
For example,
Given the following binary tree,
1
/ \
2 3
/ \ \
4 5 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ \
4-> 5 -> 7 -> NULL
# Definition for a binary tree node
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# self.next = None
class Solution:
# @param root, a tree node
# @return nothing
def connect(self, root):
if root == None:
return
queue = [root]
front = 0
rear = 1
cnt = 1
while front < rear:
fa = None
n = cnt
cnt = 0
while n > 0:
n -= 1
p = queue[front]
front += 1
if p.right != None:
queue.append(p.right)
rear += 1
cnt += 1
if p.left != None:
queue.append(p.left)
rear += 1
cnt += 1
p.next = fa
fa = pGiven a collection of numbers, return all possible permutations.
For example,[1,2,3] have the following permutations:[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2],
and [3,2,1].
class Solution {
public:
vector<vector<int> > permute(vector<int> &num) {
sort(num.begin(), num.end());
vector<vector<int> > res;
do {
res.push_back(num);
} while (next_permutation(num.begin(), num.end()));
return res;
}
};Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,[1,1,2] have the following unique permutations:[1,1,2], [1,2,1],
and [2,1,1].
class Solution {
public:
vector<vector<int> > permuteUnique(vector<int> &num) {
sort(num.begin(), num.end());
set<vector<int> > res;
do {
res.insert(num);
} while (next_permutation(num.begin(), num.end()));
return vector<vector<int> >(res.begin(), res.end());
}
};Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
For example, given s = "aab",
Return
[
["aa","b"],
["a","a","b"]
]
class Solution {
vector<vector<string>> res;
vector<string> vec;
string str;
int len;
void dfs(const int k) {
if (k == len) {
res.push_back(vec);
return;
}
for (int i = k; i < len; ++i) {
bool palindrome = true;
for (int p = k, q = i; p < q; ++p, --q) {
if (str[p] != str[q]) {
palindrome = false;
break;
}
}
if (palindrome == true) {
vec.push_back(str.substr(k, i - k + 1));
dfs(i + 1);
vec.pop_back();
}
}
}
public:
vector<vector<string>> partition(string s) {
str = s;
len = s.length();
dfs(0);
return res;
}
};Given a string s, partition s such that every substring of the partition is a palindrome.
Return the minimum cuts needed for a palindrome partitioning of s.
For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could
be produced using 1 cut.
class Solution {
public:
int minCut(string s) {
const int n = s.length();
vector<vector<bool> > isPalindrome(n, vector<bool>(n, false));
for (int i = 0; i < n; ++i) isPalindrome[i][i] = true;
for (int i = 1; i < n; ++i) isPalindrome[i - 1][i] = (s[i - 1] == s[i]);
for (int len = 3; len <= n; ++len) {
for (int i = 0, j = i + len - 1; j < n; ++i, ++j)
isPalindrome[i][j] = (s[i] == s[j] && isPalindrome[i + 1][j - 1]);
}
vector<int> dp(n + 1, INT_MAX);
dp[0] = 0;
for (int i = 0; s[i]; ++i) {
for (int j = 0; j <= i; ++j) {
if (isPalindrome[j][i])
dp[i + 1] = min(dp[j] + 1, dp[i + 1]);
}
}
return dp[n] - 1;
}
};// Precomputation: isPalindrome[N][N] in O(N^2)