Leetcode零基础学习-IDE环境(Find Anagram Mappings)

Description

Given two lists A and B, and B is an anagram of A. B is an anagram of A means B is made by randomizing the order of the elemes 

in A.We want to find an index mapping P, from A to B. A mapping P[i] = j means the i th element in A appears in B at index j.

These lists A and B may contain duplicates. If there are multiple answers, output any of them.

For example, given

A = [12, 28, 46, 32, 50]

B = [50, 12, 32, 46, 28]

We should return    (即返还的列表P中的元素为A中第0,1,2,3...个元素在B中的序号)

[1, 4, 3, 2, 0]

as P[0] = 1 because the 0 th element of A appearsat B[1], and P[1] = 4 because the 1st element of A appearsat B[4], and so on.

Note:

1.      A, B have equal lengths inrange [1, 100].

2.      A[i], B[i] are integers in range [0, 10^5].

Solution

#include
#include
#include
using namespace std;

class Solution {
public:
	vector  anagramMappings(vector &A, vector & B) {
		vector result;
		int counter = 0;
		for(int i=0;i X = { 12,28,46,32,50 };
	vector  Y = { 50,12,32,46,28 };
	vector  anagram_result;
	anagram_result = anagram.anagramMappings(X, Y);
	cout << "result= "<