每个元素指示其之前或之后的元素数量的排列?
在本节中,我们将看到一个问题。在这里,n个元素在一个数组中给出。我们必须检查该数组是否存在排列,以便每个元素指示在其之前或之后存在的元素数量。
假设数组元素为{2,1,3,3}。适当的排列类似于{3,1,2,3}。这里的前3个表示紧随其后的三个元素,1表示在此之前仅一个元素。2表示之前有两个元素,后3表示之前有三个元素。
算法
checkPermutation(arr,n)
begin define a hashmap to hold frequencies. The key and value are of integer type of the map. for each element e in arr, do increase map[e] by 1 done for i := 0 to n-1, do if map[i] is non-zero, then decrease map[i] by 1 else if map[n-i-1] is non-zero, then decrease map[n-i-1] by 1 else return false end if done return true end
示例
#include<iostream> #include<map> using namespace std; bool checkPermutation(int arr[], int n) { map<int, int> freq_map; for(int i = 0; i < n; i++){ //get the frequency of each number freq_map[arr[i]]++; } for(int i = 0; i < n; i++){ if(freq_map[i]){ //count number of elements before current element freq_map[i]--; } else if(freq_map[n-i-1]){ //count number of elements after current element freq_map[n-i-1]--; } else { return false; } } return true; } main() { int data[] = {3, 2, 3, 1}; int n = sizeof(data)/sizeof(data[0]); if(checkPermutation(data, n)){ cout << "Permutation is present"; } else { cout << "Permutation is not present"; } }
输出结果
Permutation is present