Suppose you are at a party with
n people (labeled from 0 to n - 1) and among them, there may exist one celebrity. The definition of a celebrity is that all the other n - 1people know him/her but he/she does not know any of them.
Now you want to find out who the celebrity is or verify that there is not one. The only thing you are allowed to do is to ask questions like: "Hi, A. Do you know B?" to get information of whether A knows B. You need to find out the celebrity (or verify there is not one) by asking as few questions as possible (in the asymptotic sense).
You are given a helper function
bool knows(a, b) which tells you whether A knows B. Implement a function int findCelebrity(n), your function should minimize the number of calls to knows.
Note: There will be exactly one celebrity if he/she is in the party. Return the celebrity's label if there is a celebrity in the party. If there is no celebrity, return
--------------------------------------------------------------------1.COME_BACK
对a做检测，如果a认识b || b不认识a，a就不会是celebrity
// Forward declaration of the knows API.
bool knows(int a, int b);
class Solution {
public:
int findCelebrity(int n) {
for (int a = 0; a < n; a++) {
int b = 0;
for (; b < n; b++) {
if (a == b) continue;
if (knows(a,b) || !knows(b,a)) {
break;
}
}
if (b == n) return a;
}
return -1;
}
};
第一个循坏对can进行挑选，如果i不认识can，则说明can一定不是celebrity。如果i认识can，能说明i一定不是celebrity// Forward declaration of the knows API.
bool knows(int a, int b);
class Solution {
public:
int findCelebrity(int n) {
int can = 0;
for (int i = 1; i < n; i++) {
if (!knows(i,can)) {
can = i;
}
}
for (int i = 0; i < n; i++) {
if (i == can) continue;
if (!knows(i,can) || knows(can,i)) return -1;
}
return can;
}
};
Java, 关键：
1. 所有人都认识celebrity（第二个for循环），如果谁不被人认识，那他就不是celebrity（ 第一个for循环）
2. celebrity谁也不认识（第二个for循环）
/* The knows API is defined in the parent class Relation.
boolean knows(int a, int b); */
public class Solution extends Relation {
public int findCelebrity(int n) {
int cand = 0;
for (int i = 1; i < n; i++) {
if (knows(cand, i)) {
cand = i;
}
}
for (int i = 0; i < n; i++) {
if ((cand != i && knows(cand, i)) || !knows(i, cand)) return -1;
}
return cand;
}
}
First Bad Version
You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.
Suppose you have
n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.
You are given an API
bool isBadVersion(version) which will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
------------------------------------------------------Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
Binary search
// Forward declaration of isBadVersion API.
bool isBadVersion(int version);
class Solution {
public:
int firstBadVersion(int n) {
int left = 1, right = n;
while (left <= right) {
int mid = left + (right - left) / 2;
bool bad = isBadVersion(mid);
if (bad && (mid == 1 || !isBadVersion(mid - 1))) return mid;
if (bad) {
right = mid - 1;
}else {
left = mid + 1;
}
}
return -1;
}
};
Java, recursion
/* The isBadVersion API is defined in the parent class VersionControl.
boolean isBadVersion(int version); */
public class Solution extends VersionControl {
public int firstBadVersion(int n) {
return rec(1, n);
}
private int rec(int left, int right) {
if (left > right) return left;
int mid = left + (right - left) / 2;
boolean isBad = isBadVersion(mid);
if (!isBad && isBadVersion(mid + 1)) {
return mid + 1;
}
if (isBad) {
return rec(left, mid - 1);
}
return rec(mid + 1, right);
}
}
Zigzag Iterator
Given two 1d vectors, implement an iterator to return their elements alternately.
For example, given two 1d vectors:
v1 = [1, 2] v2 = [3, 4, 5, 6]
By calling next repeatedly until hasNext returns
false, the order of elements returned by next should be: [1, 3, 2, 4, 5, 6].
Follow up: What if you are given
k 1d vectors? How well can your code be extended to such cases?
Clarification for the follow up question - Update (2015-09-18):
The "Zigzag" order is not clearly defined and is ambiguous for
The "Zigzag" order is not clearly defined and is ambiguous for
k > 2 cases. If "Zigzag" does not look right to you, replace "Zigzag" with "Cyclic". For example, given the following input:[1,2,3] [4,5,6,7] [8,9]It should return
[1,4,8,2,5,9,3,6,7].-------------------------------------------------------------
适用于k个
http://shibaili.blogspot.com/2015/08/notes-from-others.html
class ZigzagIterator {
public:
ZigzagIterator(vector<int>& v1, vector<int>& v2) {
v.push_back(v1);
v.push_back(v2);
zig = 0;
itrs = vector<int>(v.size(),0);
}
int next() {
hasNext();
int rt = v[zig][itrs[zig]];
itrs[zig]++;
zig = (zig + 1) % v.size();
return rt;
}
bool hasNext() {
for (int i = 0; i < v.size(); i++) {
if (itrs[zig] >= v[zig].size()) {
zig = (zig + 1) % v.size();
}else {
return true;
}
}
return false;
}
private:
vector<vector<int>> v;
vector<int> itrs;
int zig = 0;
};
/**
* Your ZigzagIterator object will be instantiated and called as such:
* ZigzagIterator i(v1, v2);
* while (i.hasNext()) cout << i.next();
*/
Move Zeroes
Given an array
nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.
For example, given
nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].
Note:
- You must do this in-place without making a copy of the array.
- Minimize the total number of operations.
------------------------------------------------------
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int zero = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] != 0) {
swap(nums[i],nums[zero]);
zero++;
}
}
}
};
Peeking Iterator
Given an Iterator class interface with methods:
next() and hasNext(), design and implement a PeekingIterator that support the peek() operation -- it essentially peek() at the element that will be returned by the next call to next().
Here is an example. Assume that the iterator is initialized to the beginning of the list:
[1, 2, 3].
Call
next() gets you 1, the first element in the list.
Now you call
peek() and it returns 2, the next element. Calling next() after that still return 2.
You call
next() the final time and it returns 3, the last element. Calling hasNext() after that should return false.
Hint:
- Think of "looking ahead". You want to cache the next element.
- Is one variable sufficient? Why or why not?
- Test your design with call order of
peek()beforenext()vsnext()beforepeek(). - For a clean implementation, check out Google's guava library source code.
Follow up: How would you extend your design to be generic and work with all types, not just integer?
--------------------------------------------------------------------// Below is the interface for Iterator, which is already defined for you.
// **DO NOT** modify the interface for Iterator.
class Iterator {
struct Data;
Data* data;
public:
Iterator(const vector<int>& nums);
Iterator(const Iterator& iter);
virtual ~Iterator();
// Returns the next element in the iteration.
int next();
// Returns true if the iteration has more elements.
bool hasNext() const;
};
class PeekingIterator : public Iterator {
public:
PeekingIterator(const vector<int>& nums) : Iterator(nums) {
// Initialize any member here.
// **DO NOT** save a copy of nums and manipulate it directly.
// You should only use the Iterator interface methods.
}
// Returns the next element in the iteration without advancing the iterator.
int peek() {
if (st.empty()) {
st.push(Iterator::next());
}
return st.top();
}
// hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
int next() {
if (st.empty()) return Iterator::next();
int rt = st.top();
st.pop();
return rt;
}
bool hasNext() const {
return !st.empty() || Iterator::hasNext();
}
private:
stack<int> st;
};
In Java
// Java Iterator interface reference:
// https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
class PeekingIterator implements Iterator<Integer> {
private Iterator<Integer> itr;
private Integer next;
public PeekingIterator(Iterator<Integer> iterator) {
// initialize any member here.
itr = iterator;
next = itr.hasNext() ? itr.next() : null;
}
// Returns the next element in the iteration without advancing the iterator.
public Integer peek() {
return next;
}
// hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
@Override
public Integer next() {
Integer rt = next;
next = itr.hasNext() ? itr.next() : null;
return rt;
}
@Override
public boolean hasNext() {
return next != null;
}
}
Inorder Successor in BST
Given a binary search tree and a node in it, find the in-order successor of that node in the BST.
Note: If the given node has no in-order successor in the tree, return
------------------------------------------------null.遍历
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode *mostLeft(TreeNode *root) {
while (root->left != NULL) {
root = root->left;
}
return root;
}
TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
if (p->right != NULL) {
return mostLeft(p->right);
}
TreeNode *suc = NULL;
while (root != NULL) {
if (root->val > p->val) {
suc = root;
root = root->left;
}else {
root = root->right;
}
}
return suc;
}
};
递归
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
if (root == NULL) {
return root;
}
if (root->val <= p->val) {
return inorderSuccessor(root->right,p);
}
TreeNode *next = inorderSuccessor(root->left,p);
if (next == NULL) return root;
return next;
}
};
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