33. Search in Rotated Sorted Array
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
Your algorithm’s runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
題意：給定一個旋轉有序非重複數組，一個目標值，查找該值。如果找到，返回對應座標，否則返回-1。
JAVA
class Solution {
public int search(int[] nums, int target) {
int left=0,right=nums.length-1;
while(left<=right)
{
int mid = left + ((right - left) >> 1);
if (nums[mid] == target) {
return mid;
}
if (nums[mid] < nums[right]) {
if (nums[mid] < target && target <= nums[right]) {
left = mid + 1;
}
else {
right = mid - 1;
}
}
else {
if (nums[left] <= target && target < nums[mid]) {
right = mid - 1;
}
else {
left = mid + 1;
}
}
}
return -1;
}
}
225. Implement Stack using Queues
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.
Example:
MyStack stack = new MyStack();
stack.push(1);
stack.push(2);
stack.top(); // returns 2
stack.pop(); // returns 2
stack.empty(); // returns false
用隊列來實現棧
C++
class MyStack {
public:
/** Initialize your data structure here. */
MyStack() {
}
/** Push element x onto stack. */
void push(int x) {
q.push(x);
for(int i=0;i<(int)q.size()-1;i++)
{
q.push(q.front());
q.pop();
}
}
/** Removes the element on top of the stack and returns that element. */
int pop() {
int x = q.front(); q.pop();
return x;
}
/** Get the top element. */
int top() {
return q.front();
}
/** Returns whether the stack is empty. */
bool empty() {
return q.empty();
}
private:
queue<int> q;
};
/**
* Your MyStack object will be instantiated and called as such:
* MyStack* obj = new MyStack();
* obj->push(x);
* int param_2 = obj->pop();
* int param_3 = obj->top();
* bool param_4 = obj->empty();
*/
933. Number of Recent Calls
Write a class RecentCounter to count recent requests.
It has only one method: ping(int t), where t represents some time in milliseconds.
Return the number of pings that have been made from 3000 milliseconds ago until now.
Any ping with time in [t - 3000, t] will count, including the current ping.
It is guaranteed that every call to ping uses a strictly larger value of t than before.
Example 1:
Input: inputs = [“RecentCounter”,“ping”,“ping”,“ping”,“ping”], inputs = [[],[1],[100],[3001],[3002]]
Output: [null,1,2,3,3]
寫一個RecentCounter類來計算最近的請求。
看不懂（爲了應付考試）
JAVA
class RecentCounter {
TreeSet<Integer> ts;
public RecentCounter() {
ts=new TreeSet<>();
}
public int ping(int t) {
ts.add(t);
return ts.tailSet(t-3000).size();
}
}