Given two listsAandB, andBis an anagram ofA.Bis an anagram ofAmeansBis made by randomizing the order of the elements inA.
We want to find anindex mappingP, fromAtoB. A mappingP[i] = jmeans theith element inAappears inBat indexj.
These listsAandBmay 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:
A, Bhave equal lengths in range[1, 100].
A[i], B[i]are integers in range[0, 10^5].
意思大致就是：给两个数组，数组A和数组B中有相同的数字，把数组A中每个数字在数组B中对应位置的下标输出
本来想两个for循环，遍历，但是觉得这样有可能出现TML，放弃了这种做法。
后来又想把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就ok了，执行时间7ms。
Solution：bingo
public static void main(String args[]){
int[]A = {12, 28, 46, 32, 50};
int[]B = {50, 12, 32, 46, 28};
System.out.print(anagramMappings(A,B));
}
public static int[] anagramMappings(int[] A, int[] B) {
int[]newInt = new int[A.length];
Mapbmaping = new HashMap();
for (inti=0;i
bmaping.put(B[i],i);
}
for (inti=0;i
if (bmaping.containsKey(A[i])){
//返回的是value值------>V get(Object key);
into = (int)bmaping.get(A[i]);
newInt[i] = o;
}
}
return newInt;
}
注：.get（）方法返回的是Value值
