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LeetCode 760. Find Anagram Mappi

LeetCode 760. Find Anagram Mappi

作者: 高思阳 | 来源:发表于2018-10-18 13:48 被阅读1次

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 elements in A.
We want to find an index mapping P, from A to B. A mapping P[i] = j means the ith 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
[1, 4, 3, 2, 0]
as P[0] = 1 because the 0th element of A appears at B[1], and P[1] = 4 because the 1st element of A appears at B[4], and so on.
Note:

  1. A, B have equal lengths in range [1, 100].
  2. A[i], B[i] are integers in range [0, 10^5].

题目大意:
给定数组A和B,求A中各元素在B中对应下标的数组

NSMutableArray *arrA = [NSMutableArray arrayWithArray:@[@(12), @(28), @(46), @(32), @(50)]];
    NSMutableArray *arrB = [NSMutableArray arrayWithArray:@[@(50), @(12), @(32), @(46), @(28)]];
    NSMutableDictionary *tmpDic = [NSMutableDictionary dictionary];
    
    for (NSInteger i=0; i<arrB.count; i++) {
        [tmpDic setObject:@(i) forKey:arrB[i]];
    }
    
    for (NSNumber *num in arrA) {
        NSLog(@"%@",[tmpDic objectForKey:num]);
    }

把B数组放到一个Map中(数字(key):下标(value)),遍历A数组的时候查找Map中的key,返回value

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