leetCode(Using C)——760. Find Anagram Mappings

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/lovefengruoqing/article/details/79080558

Description:

Given two lists Aand 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.

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].

Link:

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Code:

/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* anagramMappings(int* A, int ASize, int* B, int BSize, int* returnSize) {
    int *p;
    p = (int *)malloc(sizeof(int)*ASize);
    int flag[ASize];
    memset(flag, 0, sizeof(flag));

    *returnSize = ASize;
    for(int i=0; i<ASize; i++){
        for(int j=0; j<BSize; j++){
            if(A[i]==B[j]&&flag[j]==0){
                p[i]=j;
                // printf("p[%d]=%d\n",i,j);
                flag[j]=1;
                break;  //一定别忘了break！
            }
        }
    }
    return p;
}

感想

刷题一定要看清题目让你返回什么！博主一开始就写对了，但是就是在提交的时候一直结果都是空，不知道哪个地方有问题。后来仔细的研究了以下代码前面的几句说明性的话。

/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/

第一句跟你说，让你将返回的数组大小放在returnSize中。第二句让你动态的分配数组，不要直接用定义数组的方法。注意到这两点，就很简单了。