Article Directory
1. Title
A is an array [0, 1, ..., N - 1]of one arrangement, N is the length of the array A.
- Global inversion refers
i,jto meet0 <= i < j < N 并且 A[i] > A[j], - Partial inversion means that i is satisfied
0 <= i < N 并且 A[i] > A[i+1].
When the number of global inversions in array A is equal to the number of局部 inversions, return true.
示例 1:
输入: A = [1,0,2]
输出: true
解释: 有 1 个全局倒置，和 1 个局部倒置。
示例 2:
输入: A = [1,2,0]
输出: false
解释: 有 2 个全局倒置，和 1 个局部倒置。
注意:
A 是 [0, 1, ..., A.length - 1] 的一种排列
A 的长度在 [1, 5000]之间
Source: LeetCode
Link: https://leetcode-cn.com/problems/global-and-local-inversions
Copyright is owned by LeetCode . For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.
2. Problem solving
2.1 Merge sort to find the reverse order
class Solution {
int local = 0, global = 0;
vector<int> tmp;
public:
bool isIdealPermutation(vector<int>& A) {
if(A.size() <= 1) return true;
for(int i = 0; i < A.size()-1; ++i)
if(A[i] > A[i+1])
local++;
tmp.resize(A.size());
mergesort(A, 0, A.size()-1);
return local == global;//逆序度 == 局部逆序度
}
void mergesort(vector<int>& A, int l, int r)
{
if(l >= r) return;
int mid = (l+r)/2;
mergesort(A, l, mid);
mergesort(A, mid+1, r);
int i = l, j = mid+1, k = 0;
while(i <= mid && j <= r)
{
if(A[i] <= A[j])
tmp[k++] = A[i++];
else//后序写入时，检查前面没有出队的（比我大，我在后面）
{
tmp[k++] = A[j++];
global += mid-i+1;
}
}
while(i <= mid)
tmp[k++] = A[i++];
while(j <= r)
tmp[k++] = A[j++];
k = 0; i = l;
while(i <= r)
A[i++] = tmp[k++];
}
};
Time complexity O (n log n) O(n\log n)O ( nlogn )
192 ms 33.8 MB
2.2 Binary search
- The number of global inversions must be >= the number of local inversions
- Check each number A[i] A[i]A [ i ] , before it[0, i − 2] [0, i-2][0,i−2 ] (Neighboring does not need to be checked) There is a larger number, which certainly does not satisfy the meaning of the question
class Solution {
public:
bool isIdealPermutation(vector<int>& A) {
set<int> s;
s.insert(A[0]);
for(int i = 2; i < A.size(); ++i)
{
if(s.lower_bound(A[i]+1) != s.end())
{
//前面存在比 A[i] 大的数
return false;
}
s.insert(A[i-1]);
}
return true;
}
};
Time complexity O (n log n) O(n\log n)O ( nlogn )
396 ms 52.1 MB
2.3 One traversal
Optimization: On the basis of the second method, record the previous maximum value
class Solution {
public:
bool isIdealPermutation(vector<int>& A) {
int MAX = A[0];
for(int i = 2; i < A.size(); ++i)
{
if(MAX > A[i])
return false;
MAX = max(MAX, A[i-1]);
}
return true;
}
};
Time complexity O (n) O(n)O ( n )
148 ms 32.5 MB
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