如何解决最长递增子序列Onlogn
| LIS：维基百科 我不明白一件事： 为什么X [M [i]]是不递减的序列？解决方法
首先看一下n ^ 2算法：
dp[0] = 1;
for( int i = 1; i < len; i++ ) {
dp[i] = 1;
for( int j = 0; j < i; j++ ) {
if( array[i] > array[j] ) {
if( dp[i] < dp[j]+1 ) {
dp[i] = dp[j]+1;
}
}
}
}
sz = 1;
c[1] = array[0]; /*at this point,the minimum value of the last element of the size 1 increasing sequence must be array[0]*/
dp[0] = 1;
for( int i = 1; i < len; i++ ) {
if( array[i] < c[1] ) {
c[1] = array[i]; /*you have to update the minimum value right now*/
dp[i] = 1;
}
else if( array[i] > c[sz] ) {
c[sz+1] = array[i];
dp[i] = sz+1;
sz++;
}
else {
int k = binary_search( c,sz,array[i] ); /*you want to find k so that c[k-1]<array[i]<c[k]*/
c[k] = array[i];
dp[i] = k;
}
}
set<int> st;
set<int>::iterator it;
st.clear();
for(i=0; i<n; i++) {
st.insert(array[i]);
it=st.find(array[i]);
it++;
if(it!=st.end()) st.erase(it);
}
cout<<st.size()<<endl;
set<int> st;
set<int>::iterator it;
st.clear();
for(int i=0; i<n; i++) {
it = st.lower_bound(a[i]);
if (it != st.end()) st.erase(it);
st.insert(a[i]);
}
cout<<st.size()<<endl;
typedef pair<int,int> IndexValue;
struct IndexValueCompare{
inline bool operator() (const IndexValue &one,const IndexValue &another){
return one.second < another.second;
}
};
vector<int> LIS(const vector<int> &sequence){
vector<int> parent(sequence.size());
set<IndexValue,IndexValueCompare> s;
for(int i = 0; i < sequence.size(); ++i){
IndexValue iv(i,sequence[i]);
if(i == 0){
s.insert(iv);
continue;
}
auto index = s.lower_bound(iv);
if(index != s.end()){
if(sequence[i] < sequence[index->first]){
if(index != s.begin()) {
parent[i] = (--index)->first;
index++;
}
s.erase(index);
}
} else{
parent[i] = s.rbegin()->first;
}
s.insert(iv);
}
vector<int> result(s.size());
int index = s.rbegin()->first;
for(auto iter = s.rbegin(); iter != s.rend(); index = parent[index],++iter){
result[distance(iter,s.rend()) - 1] = sequence[index];
}
return result;
}
k
k+1
set<int> my_set;
set<int>::iterator it;
vector <int> out;
out.clear();
my_set.clear();
for(int i = 1; i <= n; i++) {
my_set.insert(a[i]);
it = my_set.find(a[i]);
it++;
if(it != my_set.end())
st.erase(it);
else
out.push_back(*it);
}
cout<< out.size();
//! card piles contain pile of cards,nth pile contains n cards.
int top_card_list[n+1];
for(int i = 0; i <= n; i++) {
top_card_list[i] = -1;
}
top_card_list
n
3
* 7 2
-------------------------------------------------------------
Pile of cards above (top card is larger than lower cards)
(note that pile of card represents longest increasing subsequence too !)
O(n)
for(int i = 1; i < n; i++) { // outer loop
for(int j = 0; j < i; j++) { // inner loop
if(arr[i] > arr[j]) {
if(memo_len[i] < (memo_len[j]+1)) {
// relaxation
memo_len[i] = memo_len[j]+1;
result = std::max(result,memo_len[i]);
pred[i] = j;
}
}
}
}
O(number of piles)
O(number of cards)
for(int i = 1; i < n; i++) {
pile_height[i] = 1;
const int j = pile_search(top_card_list,arr,pile_len,arr[i]);
if(arr[i] > arr[j]) {
if(pile_height[i] < (pile_height[j]+1)) {
// relaxation
pile_height[i] = pile_height[j]+1;
result = std::max(result,pile_height[i]);
pile_len = std::max(pile_len,pile_height[i]);
}
}
if(-1 == top_card_list[pile_height[i]] || arr[top_card_list[pile_height[i]]] > arr[i]) {
top_card_list[pile_height[i]] = i; // top card on the pile is now i
}
}
inline static int pile_search(const int*top_card_list,const vector<int>& arr,int pile_len,int strict_upper_limit) {
int start = 1,bound=pile_len;
while(start < bound) {
if(arr[top_card_list[bound]] < strict_upper_limit) {
return top_card_list[bound];
}
int mid = (start+bound)/2 + ((start+bound)&1);
if(arr[top_card_list[mid]] >= strict_upper_limit) {
// go lower
bound = mid-1;
} else {
start = mid;
}
}
return top_card_list[bound];
}
top_card_list[]
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