剑指offer——面试题51：数组中重复的数字

力扣

法1：使用HashSet

Python

class Solution:
    def findRepeatDocument(self, documents: List[int]) -> int:
        h = set()
        for item in documents:
            if item in h:
                return item
            else:
                h.add(item)

Java

    public int findRepeatNumber(int[] array) {
        Map<Integer, Integer> numToCountMap = new HashMap<>();
        for (int num : array) {
            if (numToCountMap.containsKey(num)) {
                return num;
            } else {
                numToCountMap.put(num, 1);
            }
        }
        return -1;
    }

法2：使用索引=>值映射

虽然不是很好理解。。

class Solution {
    public int findRepeatDocument(int[] documents) {
        int i = 0;
        while(i < documents.length) {
            if(documents[i] == i) {
                i++;
                continue;
            }
            if(documents[documents[i]] == documents[i]) return documents[i];
            int tmp = documents[i];
            documents[i] = documents[tmp];
            documents[tmp] = tmp;
        }
        return -1;
    }
}

##Solution1:
20180910更新。利用数组做一次hash映射，时间复杂度为$O(n)$，空间复杂度$O(n)$.

class Solution {
public:
    // Parameters:
    //        numbers:     an array of integers
    //        length:      the length of array numbers
    //        duplication: (Output) the duplicated number in the array number
    // Return value:       true if the input is valid, and there are some duplications in the array number
    //                     otherwise false
    bool duplicate(int numbers[], int length, int* duplication) {
        int hash[length];
        memset(hash, 0, sizeof(hash));
        for (int i = 0; i < length; i++) {
            hash[numbers[i]]++;
            if (hash[numbers[i]] >= 2) {
                *duplication = numbers[i];
                return true;
            }
        }
        return false;
    }
};

##Solution2:
参考网址：https://www.nowcoder.com/profile/4747275/codeBookDetail?submissionId=16801123
思路：不需要额外的数组或者hash table来保存，题目里写了数组里数字的范围保证在0 ~ n-1 之间，所以可以利用现有数组设置标志，当一个数字被访问过后，可以设置对应位上的数 + n，之后再遇到相同的数时，会发现对应位上的数已经大于等于n了，那么直接返回这个数即可。
点评：对于此类问题很好的思路。值得特殊记一下！
时间复杂度$O(n)$，空间复杂度$O(1)$.
20180910晚：思路确实巧妙啊~~~

class Solution {
public:
    // Parameters:
    //        numbers:     an array of integers
    //        length:      the length of array numbers
    //        duplication: (Output) the duplicated number in the array number
    // Return value:       true if the input is valid, and there are some duplications in the array number
    //                     otherwise false
    bool duplicate(int numbers[], int length, int* duplication) {
        int index;
        for(int i = 0; i < length; i++) {
            index = numbers[i];
            if(index >= length) {
                index -= length;
            }
            if(numbers[index] > length) {
                *duplication = index;
                return true;
            }
            numbers[index] += length;
        }
        return false;
    }
};