int find_kth(vector<int>::iterator nums1, int m, vector<int>::iterator nums2, int n, int k){
//always let nums1's size smaller or equal to nums2
if(m>n){
return find_kth(nums2,n,nums1,m,k);
}
if(m==0){
return *(nums2+k-1);
}
if(k==1){
return min(*nums1,*nums2);
}
//divide k into tow parts
int mid1=min(k/2,m),mid2=k-mid1;
vector<int>::iterator iter;
if(*(nums1+mid1-1)<*(nums2+mid2-1)){
return find_kth(nums1+mid1,m-mid1,nums2,n,k-mid1);
}else if (*(nums1+mid1-1)>*(nums2+mid2-1)){
return find_kth(nums1,m,nums2+mid2,n-mid2,k-mid2);
}else{
return nums1[mid1-1];
}
}
double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
int total = nums1Size + nums2Size;
if(total & 0x1)
return find_kth(nums1,nums1Size,nums2,nums2Size,total/2+1);
else
return (find_kth(nums1,nums1Size,nums2,nums2Size,total/2)+find_kth(nums1,nums1Size,nums2,nums2Size,total/2+1))/2.0;
}
double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
int all = nums1Size + nums2Size;
int k = all / 2 + 1;
double ans = find_kth(nums1, nums1Size, nums2, nums2Size, k);
if(all % 2)
return ans;
return (ans + find_kth(nums1, nums1Size, nums2, nums2Size, k-1)) / 2;
}
double findMedianSortedArrays(std::vector<int>& nums1, std::vector<int>& nums2)
{
const int total = nums1.size() + nums2.size();
if (total % 2)//奇数
return find_kth(nums1, 0, nums2, 0, total / 2 + 1);
else//偶数
return (find_kth(nums1, 0, nums2, 0, total / 2) + find_kth(nums1, 0, nums2, 0, total / 2 + 1)) / 2;
}
double findMedianSortedArrays(int A[], int m, int B[], int n) {
int total = m + n;
if(total & 0x1) { //奇数个元素,取中间元素
return find_kth(A, m, B, n, total/2 + 1);
}
else { //偶数个元素,取中间两个元素的平均
return (find_kth(A, m, B, n, total/2) + find_kth(A, m, B, n, total/2 + 1))/2;
}
}
// There are two sorted arrays nums1 and nums2 of size m and n respectively.
// Find the median of the two sorted arrays.
// The overall run time complexity should be O(log (m+n)).
// The median: https://en.wikipedia.org/wiki/Median
double Array::findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2)
{
const int m = nums1.size(), n = nums2.size();
const int total = m + n;
if (total & 1)
return find_kth(nums1.begin(), m, nums2.begin(), n, total / 2 + 1);
else
return (find_kth(nums1.begin(), m, nums2.begin(), n, total / 2)
+ find_kth(nums1.begin(), m, nums2.begin(), n, total / 2 + 1)) / 2.0;
}
int main() {
List l;
for (int i = 0; i < 10; i++)
l.add(i);
assert(find_kth(l, 0) == 9);
assert(find_kth(l, 1) == 8);
assert(find_kth(l, 9) == 0);
}
TreapNode * find_kth(TreapNode*x, int k) {
if (x == NULL) return NULL;
int lsz = LSZ(*x);
if (k <= lsz) {
return find_kth(x->ch[0], k);
} else if (k == lsz + 1) {
return x;
} else {
return find_kth(x->ch[1], k - lsz - 1);
}
}
/*
Divide and Conquer inspired by find k-th number in sorted array.
The complexity is of course O(log(M+N)).
Similiar with the following answer except without slicing.
https://discuss.leetcode.com/topic/6947/intuitive-python-o-log-m-n-solution-by-kth-smallest-in-the-two-sorted-arrays-252ms
*/
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
int len1 = nums1.size();
int len2 = nums2.size();
int len = len1 + len2;
if(len & 0x1 == 1){
return find_kth(nums1, 0, len1, nums2, 0, len2, (len+1)/2);
}
else{
return (find_kth(nums1, 0, len1, nums2, 0, len2, (len)/2) +
find_kth(nums1, 0, len1, nums2, 0, len2, (len)/2+1)) / 2.0;
}
}
void find_kth(vector<int>& a, int left, int right, int k)
{
int pivot = a[right];
int pos = left;
for (int i=left;i<right;++i)
if (a[i] < a[right])
swap(a[i], a[pos++]);
swap(a[pos],a[right]);
if (k == pos) return;
if (pos < k) find_kth(a, pos+1, right, k);
if (pos > k) find_kth(a, left, pos-1, k);
}
int find_kth(int *A, int m, int *B, int n, int k){//查找第k小的值
if(m>n) return find_kth(B,n,A,m,k); //保证m<=n
if(m==0) return B[k-1]; //终止条件1：其中一个长度为0；
if(k==1) return min(A[0],B[0]);//终止条件2：查找最小值；
int ia=min(k/2,m),ib=k-ia;
if(A[ia-1]<B[ib-1]) //若A第ia个元素小，抛弃左侧元素
return find_kth(A+ia,m-ia,B,n,k-ia);
else if(A[ia-1]>B[ib-1]) //若B第ib个元素小，抛弃左侧元素
return find_kth(A,m,B+ib,n-ib,k-ib);
else
return A[ia-1]; //终止条件3：找到了目标值；
}
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
int m=nums1.size();
int n=nums2.size();
int total = m+n;
if(total & 0x1){
return find_kth(nums1.begin(),m,nums2.begin(),n,total/2+1);
}else{
return (find_kth(nums1.begin(),m,nums2.begin(),n,total/2)
+find_kth(nums1.begin(),m,nums2.begin(),n,total/2+1))/2.0;
}
}
int Array::find_kth(vector<int>::const_iterator a, int m, vector<int>::const_iterator b, int n, int k)
{
if (m > n) return find_kth(b, n, a, m, k);
if (m == 0) return *(b + k - 1);
if (k == 1) return min(*a, *b);
// divide k into two parts
int ia = min(k / 2, m), ib = k - ia;
if (*(a + ia - 1) < *(b + ib - 1))
return find_kth(a + ia, m - ia, b, n, k - ia);
else if (*(a + ia - 1) > *(b + ib - 1))
return find_kth(a, m, b + ib, n - ib, k - ib);
else
return a[ia - 1];
}
int find_kth(int a[],int b[],int size_a,int size_b,int k)
{
if(size_a > size_b) return find_kth(b,a,size_b,size_a,k);
// make the smaller array as A
if(size_a == 0 && size_b > 0) return b[k-1];
// if one array is empty simply return kth index value
if(k==1) return (a[0]<=b[0])? a[0]:b[0];
// if k is the smallest element return smallest value
int i = (size_a <= k/2)? size_a: k/2;
int j = (size_b <= k/2)? size_b: k/2;
//find the k/2 index in the array or the max length if less
if(a[i-1] < b[j-1]) return find_kth(a+i,b,(size_a-i),j,(k-i));
else return find_kth(a,b+j,i,(size_b-j),(k-j));
//discard the side of both the arrays in which k cannot be present
}
double find_kth(int * a, int n1, int * b, int n2, int k){
if(n1 == 0)
return b[k-1];
if(n2 == 0)
return a[k-1];
if(k == 1)
return a[0]<b[0]?a[0]:b[0];
int amid = n1 / 2;
int bmid = n2 / 2;
if(a[amid] <= b[bmid]){
if(amid + bmid + 1 >= k)
return find_kth(a, n1, b, bmid, k);
return find_kth(a+amid+1, n1-amid-1, b, n2, k-amid-1);
}
return find_kth(b, n2, a, n1, k);
}
void main(void)
{
int a[100];
int b[100];
int size_a = 0;
int size_b = 0;
while(1)
{
int var = 0;
printf("Enter Array A elements (end with -1)\n");
scanf("%d",&var);
if(var == -1) break;
a[size_a++] = var;
}
while(1)
{
int var = 0;
printf("Enter Array B elements (end with -1)\n");
scanf("%d",&var);
if(var == -1) break;
b[size_b++] = var;
}
printf("ENTER THE VALUE FOR K\n");
int k;
scanf("%d",&k);
int answer = find_kth(a,b,size_a,size_b,k);
printf("The %dth smallest element is %d\n",k,answer);
}
double find_kth(int A[], int m, int B[], int n, int k) {
if(m > n) return find_kth(B, n, A, m, k); //假定m小于等于n,这样处理更方便
//按照前面的分析，下面是两个终止的情况
if(m == 0) return B[k-1];
if(k == 1) return min(A[0], B[0]);
//pa取k/2，如果m比k/2小，pa取m
int pa = min(k/2, m), pb = k - pa; //二分法的变形
if(A[pa-1] < B[pb-1]) {
//这种情况可以放心去掉A的pa个元素（k/2或者全部）
return find_kth(A+pa, m-pa, B, n, k-pa);
}
else if(A[pa-1] > B[pb-1]) {
//这种情况可以放心去掉B的pb个元素（k/2或者更多）
return find_kth(A, m, B+pb, n-pb, k-pb);
}
else {
return A[pa-1];
}
}
void left(int ax,int ay)
{
int p,i,f;
for(i=ax-r;i<=ax+r;i++)
{
plus(old[i][ay-r-1],-1);
plus(old[i][ay+r],1);
}
p=find_kth(k);
neww[ax][ay]=p;
}
double find_kth(std::vector<int>& A, int m, std::vector<int>& B, int n, int k)
{
if (A.size() > B.size())//默认A数组要小
return find_kth(B, n, A, m, k);
if (m >= A.size())//A数组为空，说明第K小的元素在数组B中
return B[n + k - 1];
if (n >= B.size())
return A[m + k - 1];
if (k == 1)//找第1小的元素
return std::min(A[m], B[n]);
int pa = std::min(k / 2, int(A.size() - m)), pb = k - pa;//注意pb的算法，k-pa可以使pa和pb加起来等于k
if (A[m + pa - 1] < B[n + pb - 1])//如果小于，表示数组A的pa-1位置以下的元素都不是，可以删除
return find_kth(A, m + pa, B, n, k - pa);
else if (A[m + pa - 1] > B[n + pb - 1])//如果大于，表示数组B的pb - 1下标以下的元素都不是第k小的元素，可以删除
return find_kth(A, m, B, n + pb, k - pb);
else//相等，表示这就是第k小的元素，因为pa+pb等于k
return A[m + pa - 1];
}
void up(int ax,int ay)
{
int p,i,f;
for(f=ay-r;f<=ay+r;f++)
{
plus(old[ax-r-1][f],-1);
plus(old[ax+r][f],1);
}
p=find_kth(k);
neww[ax][ay]=p;
}
int find_kth(vector<int>& list1, int begin1, int end1,
vector<int>& list2, int begin2, int end2, int k){
/*
Find the kth number in two sorted list: list1 , list2
Binary search as followers:
Firstly cut list1 and list2 into two parts by t1 and t2, respectively.
1. lis1_left ... list1[t1-th] ... list1_right,
2. lis2_left ... list2[t2-th] ... list2_right
Then compare value of list1[t1-th] and list2[t2-th] in list2.
Three situations about the relation between list1[t1-th] and list2[t2-th]:
1. < Equal the (k-t1)th number in list1_right and list_2 left.
2. > Equal the (k-t2)th number in list1_left and list_2 right.
3. == Find the k-th number.
*/
int len1 = end1 - begin1;
int len2 = end2 - begin2;
if(len1 > len2){
return find_kth(list2, begin2, end2, list1, begin1, end1, k);
}
if(len1 == 0){
return list2[begin2+k-1];
}
if(k==1){
return min(list1[begin1], list2[begin2]);
}
int t1 = min(k/2, len1);
int t2 = k - t1;
if(list1[begin1+t1-1] < list2[begin2+t2-1]){
return find_kth(list1, begin1+t1, end1, list2, begin2, begin2+t2, k-t1);
}
else if(list1[begin1+t1-1] > list2[begin2+t2-1]){
return find_kth(list1, begin1, begin1+t1, list2, begin2+t2, end2, k-t2);
}
else{
return list1[begin1+t1-1];
}
}
void fun()
{
int ax,ay;
int i,f,p;
for(i=1;i<=2*r+1;i++)
{
for(f=1;f<=2*r+1;f++)
{
plus(old[i][f],1);
}
}
p=find_kth(k);
neww[r+1][r+1]=p;
for(ax=r+1;ax<=n-r;ax++)
{
if(1 == 1&(ax-r))
{
for(ay=r+1;ay<=n-r;ay++)
{
if(ax==ay&&ax==r+1)
{
continue;
}
if(ay == r+1)
{
up(ax,ay);
}
else
{
left(ax,ay);
}
}
}
else
{
for(ay=n-r;ay>=r+1;ay--)
{
if(ay == n-r)
{
up(ax,ay);
}
else
{
right(ax,ay);
}
}
}
}
}
void solve()
{
value[++total] = -99999999;
count[total] = 1;
root = total;
int N;
std::scanf("%d", &N);
for(int i = 0; i != N; ++i)
{
int opt, x;
std::scanf("%d %d", &opt, &x);
switch(opt)
{
case 1:
insert(x);
break;
case 2:
remove(x);
break;
case 3:
std::printf("%d\n", find_rank(x) - 1);
break;
case 4:
std::printf("%d\n", find_kth(x + 1));
break;
case 5:
opt = find_prev(x);
splay(opt);
std::printf("%d\n", value[opt]);
break;
case 6:
opt = find_next(x);
splay(opt);
std::printf("%d\n", value[opt]);
break;
}
}
}
void wiggleSort(vector<int>& a)
{
if (a.size()<2) return ;
// deal with odd length
int len = a.size();
if (len % 2 == 1)
{
int min_pos = 0;
for (int i=1;i<len;++i)
if (a[i] < a[min_pos]) min_pos = i;
swap(a[min_pos], a[--len]);
}
int k = len/2 - 1;
find_kth(a, 0, len-1, k);
// a[k] will be the kth number in a
// a[0]..a[k-1] are all leq a[k]
// a[k+1]..a[len-1] are all geq a[k]
// cout << k << endl;
// for (int i=0;i<a.size();++i)
// cout << a[i] << " ";
// cout << endl;
int special = a[k];
// cout << "special: " << special << endl;
for (int i=1;i<=k;i+=2)
swap(a[i],a[k+1+i]);
for (int i=1;i<len;i+=2)
if (a[i-1] > a[i]) swap(a[i-1],a[i]);
// cout << "Make paris" << endl;
// for (int i=0;i<a.size();++i)
// cout << a[i] << " ";
// cout << endl;
int p1 = 0, p2 = 0;
for (;p2<len;p2+=2)
if (a[p2] == a[p2+1])
{
while (p1<len && (a[p1]==special ||a[p1+1] == special)) p1+=2;
swap(a[p1],a[p2]);
}
// cout << "Solve conflicts within a pair" << endl;
// for (int i=0;i<a.size();++i)
// cout << a[i] << " ";
// cout << endl;
p1 = 0;
for (p2=0;p2<len;p2+=2)
if (a[p2] == special)
{
while (p1<len && a[p1] == special) p1 += 2;
if (p1 >= p2) continue;
swap(a[p1],a[p2]);
swap(a[p1+1],a[p2+1]);
}
// cout << "Solve conflicts between pairs" << endl;
// for (int i=0;i<a.size();++i)
// cout << a[i] << " ";
// cout << endl;
}
int find_kth(int k) {
if (k < 1 || k > size()) return -1;
TreapNode *x = find_kth(root, k);
if (x) return x->val + delta;
return -1;
}