3. 数组中重复的数字

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题目描述

题目：在一个长度为n的数组里的所有数字都在0到n-1的范围内。数组中某些数字是重复的，但不知道有几个数字重复了，也不知道每个数字重复了几次。请找出数组中任意一个重复的数字。例如，如果输入长度为7的数组{2, 3, 1, 0, 2, 5, 3}，那么对应的输出是重复的数字2或者3。

题目链接

https://www.nowcoder.com/practice/623a5ac0ea5b4e5f95552655361ae0a8?tpId=13&tqId=11203&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking

解题思路

https://github.com/CyC2018/CS-Notes/blob/master/notes/%E5%89%91%E6%8C%87%20offer%20%E9%A2%98%E8%A7%A3.md#4-%E4%BA%8C%E7%BB%B4%E6%95%B0%E7%BB%84%E4%B8%AD%E7%9A%84%E6%9F%A5%E6%89%BE

代码

C++ Code

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) {
        if(numbers == nullptr || length <= 0)
            return false;
        for(int i = 0; i < length; i++)
        {
            if(numbers[i] < 0 || numbers[i] > length - 1)
                return false;
        }

        for(int i = 0; i < length; i++)
        {
            while(numbers[i] != i)
            {
                if(numbers[i] == numbers[numbers[i]])
                {
                    *duplication = numbers[i];
                    return true;
                }

                int temp = numbers[i];
                numbers[i] = numbers[temp];
                numbers[temp] = temp;
            }
        }
        return false;
    }
};

Java Code

public class Solution {
    // Parameters:
    //    numbers:     an array of integers
    //    length:      the length of array numbers
    //    duplication: (Output) the duplicated number in the array number,length of duplication array is 1,so using duplication[0] = ? in implementation;
    //                  Here duplication like pointor in C/C++, duplication[0] equal *duplication in C/C++
    //    这里要特别注意~返回任意重复的一个，赋值duplication[0]
    // Return value:       true if the input is valid, and there are some duplications in the array number
    //                     otherwise false
    public boolean duplicate(int numbers[],int length,int [] duplication) {
        if (numbers == null || length <= 0)
            return false;
        for (int i = 0; i < length; i++){
            while (numbers[i] != i){
                if (numbers[i] == numbers[numbers[i]]){
                    duplication[0] = numbers[i];
                    return true;
                }
                swap(numbers, i, numbers[i]);
            }
        }
        return false;
    }

    private void swap(int[] numbers, int i, int j){
        int t = numbers[i];
        numbers[i] = numbers[j];
        numbers[j] = t;
    }
}

Python Code

# -*- coding:utf-8 -*-
class Solution:
    # 这里要特别注意~找到任意重复的一个值并赋值到duplication[0]
    # 函数返回True/False
    def duplicate(self, numbers, duplication):
        # write code here
        if not isinstance(numbers, list) or len(numbers) <= 0:
            return False

        for num in numbers:
            if not isinstance(num, int) or num >= len(numbers) or num < 0:
                return False

        for index in range(len(numbers)):
            while numbers[index] != index:
                if numbers[numbers[index]] == numbers[index]:
                    # 找到重复数字

                    duplication[0] = numbers[index]
                    return True
                numbers[numbers[index]], numbers[index] = numbers[index], numbers[numbers[index]]

        return False