You are given two?0-indexed?integer?permutations?A?and?B?of length?n.

A?prefix common array?of?A?and?B?is an array?C?such that?C[i]?is equal to the count of numbers that are present at or before the index?i?in both?A?and?B.

Return?the?prefix common array?of?A?and?B.

A sequence of?n?integers is called a?permutation?if it contains all integers from?1?to?n?exactly once.

?

Example 1:

Input: A = [1,3,2,4], B = [3,1,2,4]

Output: [0,2,3,4]

Explanation:?

At i = 0: no number is common, so C[0] = 0.?

At i = 1: 1 and 3 are common in A and B, so C[1] = 2.?

At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.?

At i = 3: 1, 2, 3, and 4 are common in A and B, so C[3] = 4.

Example 2:

Input: A = [2,3,1], B = [3,1,2]

Output: [0,1,3]

Explanation:?

At i = 0: no number is common, so C[0] = 0.?

At i = 1: only 3 is common in A and B, so C[1] = 1.?

At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.

?

Constraints:

• 1 <= A.length == B.length == n <= 50

• 1 <= A[i], B[i] <= n

• It is guaranteed that A and B are both a permutation of n integers.

判斷前i個(gè)位置有相同的數(shù)字的個(gè)數(shù)；

這里面用一個(gè)集合去存儲兩個(gè)數(shù)組前i個(gè)數(shù)，當(dāng)下一個(gè)出線新的數(shù)字的時(shí)候，如果存在就++；

當(dāng)然如果2個(gè)相等，直接++即可；

所有可能性都想到了，就返回即可；

一開始是想用的笨的方法；以后還是多想想快速的解決方案了。。。

Runtime:?5 ms, faster than?77.64%?of?Java?online submissions for?Find the Prefix Common Array of Two Arrays.

Memory Usage:?43.7 MB, less than?15.92%?of?Java?online submissions for?Find the Prefix Common Array of Two Arrays.