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Copy path775. Global and Local Inversions.cpp
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775. Global and Local Inversions.cpp
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We have some permutation A of [0, 1, ..., N - 1], where N is the length of A.
The number of (global) inversions is the number of i < j with 0 <= i < j < N and A[i] > A[j].
The number of local inversions is the number of i with 0 <= i < N and A[i] > A[i+1].
Return true if and only if the number of global inversions is equal to the number of local inversions.
Example 1:
Input: A = [1,0,2]
Output: true
Explanation: There is 1 global inversion, and 1 local inversion.
Example 2:
Input: A = [1,2,0]
Output: false
Explanation: There are 2 global inversions, and 1 local inversion.
Note:
A will be a permutation of [0, 1, ..., A.length - 1].
A will have length in range [1, 5000].
The time limit for this problem has been reduced.
class Solution {
public:
int localInv=0, globalInv=0;
void merge(vector<int> &v, int l, int mid, int r){
int n1=mid-l+1;
int n2=r-(mid+1)+1;
vector<int> L(n1);
vector<int> R(n2);
for(int i=0;i<n1;i++)
L[i]=v[l+i];
for(int j=0;j<n2;j++)
R[j]=v[mid+1+j];
int i=0, j=0, k=l;
while(i<n1 && j<n2){
if(L[i]<=R[j]){
v[k]=L[i];
i++;
} else {
v[k]=R[j];
j++;
globalInv+=n1-i;
}
k++;
}
while(i<n1){
v[k]=L[i];
i++;
k++;
}
while(j<n2){
v[k]=R[j];
j++;
k++;
}
}
void mergeSort(vector<int> &v, int l, int r){
if(l<r){
int mid=l+(r-l)/2;
mergeSort(v, l, mid);
mergeSort(v, mid+1, r);
merge(v, l, mid, r);
}
}
bool isIdealPermutation(vector<int>& v) {
// cal local inv
int n=v.size();
for(int i=0;i<n-1;i++){
if(v[i]>v[i+1]) localInv++;
}
// cal global inv
mergeSort(v, 0, n-1);
if(localInv==globalInv) return true;
else return false;
}
};