Question Definition
Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.
For example,
Given [3,2,1,5,6,4] and k = 2, return 5.
Note:
You may assume k is always valid, 1 ≤ k ≤ array’s length.
Java Solution
class Solution {
public void wiggleSort(int[] nums) {
int median = findKthLargest(nums, (nums.length + 1) / 2);
int n = nums.length;
int left = 0, i = 0, right = n - 1;
while (i <= right) {
if (nums[newIndex(i,n)] > median) {
swap(nums, newIndex(left++,n), newIndex(i++,n));
}
else if (nums[newIndex(i,n)] < median) {
swap(nums, newIndex(right--,n), newIndex(i,n));
}
else {
i++;
}
}
}
private void swap(int[] nums,int i,int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
private int newIndex(int index, int n) {
return (1 + 2*index) % (n | 1);
}
private int findKthLargest(int[] nums, int k) {
int start = 0;
int end = nums.length - 1;
while(true){
int partition = partition(nums, start, end);
if(partition == k - 1)
return nums[partition];
else if(partition > k - 1){
end = partition - 1;
}else{
start = partition + 1;
}
}
}
private int partition(int[] a, int lo, int hi){
int i = lo, j = hi+1; // left and right scan indices
int v = a[lo]; // partitioning item
while (true)
{ // Scan right, scan left, check for scan complete, and exchange.
while (++i < hi && a[i] > v);
while (--j > lo && v > a[j]) if (j == lo) break;
if (i >= j) break;
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
int temp = a[lo];
a[lo] = a[j];
a[j] = temp;
return j;
}
}
Question Definition
You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k.
Define a pair (u,v) which consists of one element from the first array and one element from the second array.
Find the k pairs (u1,v1),(u2,v2) …(uk,vk) with the smallest sums.
Example 1:
Given nums1 = [1,7,11], nums2 = [2,4,6], k = 3
Return: [1,2],[1,4],[1,6]
The first 3 pairs are returned from the sequence:
[1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]
Example 2:
Given nums1 = [1,1,2], nums2 = [1,2,3], k = 2
Return: [1,1],[1,1]
The first 2 pairs are returned from the sequence:
[1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]
Example 3:
Given nums1 = [1,2], nums2 = [3], k = 3
Return: [1,3],[2,3]
All possible pairs are returned from the sequence:
[1,3],[2,3]
Java Solution
public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
PriorityQueue<int[]> que = new PriorityQueue<>((a,b)->a[0]+a[1]-b[0]-b[1]);
List<int[]> res = new ArrayList<>();
if(nums1.length==0 || nums2.length==0 || k==0) return res;
for(int i=0; i<nums1.length && i<k; i++) que.offer(new int[]{nums1[i], nums2[0], 0});
while(k-- > 0 && !que.isEmpty()){
int[] cur = que.poll();
res.add(new int[]{cur[0], cur[1]});
if(cur[2] == nums2.length-1) continue;
que.offer(new int[]{cur[0],nums2[cur[2]+1], cur[2]+1});
}
return res;
}
Question Definition
Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.
More …
Question Definition
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
For 1-byte character, the first bit is a 0, followed by its unicode code.
For n-bytes character, the first n-bits are all one’s, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
This is how the UTF-8 encoding would work:
Char. number range | UTF-8 octet sequence
(hexadecimal) | (binary)
——————–+———————————————
0000 0000-0000 007F | 0xxxxxxx
0000 0080-0000 07FF | 110xxxxx 10xxxxxx
0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
Given an array of integers representing the data, return whether it is a valid utf-8 encoding.
More …
Question Definition
Given a linked list, swap every two adjacent nodes and return its head.
For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.
Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
Java Solution
public ListNode swapPairs(ListNode head) {
ListNode cur = head;
while(head != null && head.next != null){
int temp = head.val;
head.val = head.next.val;
head.next.val = temp;
head = head.next.next;
}
return cur;
}